11

Riemann curvature and the Ricci tensor

11.1 What is curvature?
11.2 Tidal forces
11.3 Riemann curvature 121
11.4 Symmetries of the Riemann tensor of Rieman
11.5 The Ricci tensor and Ricci scalar
11.6 Example computations 128 11.7 Geodesic deviation revisited
Chapter summary 130
Exercises
1 1 ^(1){ }^{1}1 Our problem in finding the curvature of spacetime is more acute, of course, since we can't use any method analogous to noting the change in elevation of distant stars to work out the nature of the curvature.
2 2 ^(2){ }^{2}2 G. F. Bernhard Riemann (1826-1866)
3 3 ^(3){ }^{3}3 The motion of particles in curved spacetime is often described in terms of a rubber sheet model. This involves picturing space (in two dimensions) as being akin to an elastic sheet. Massive objects make indentations in the sheet and so all objects have their motion affected by having to negotiate these indentations as they move around. Although useful as a picture, it should also be kept in mind that this model describes the curvature of space and not spacetime. In some cases (such as the precession of the orbit of a planet around a star), the metric component describing the timelike variable has a larger effect on the corrections to Newtonian motion than those relating to the spatial coordinates. The combined effect is therefore certainly a spacetim one. This is another manifestation one. This is another manifestation of the feature that space and time are inextricably linked in relativity.
From the point of view of general relativity, humans are condemned to be trapped in spacetime. We cannot, for example, lift ourselves out of spacetime and look at its structure from the outside, by embedding it in some higher dimensional space. If gravitation is manifested in terms of the curvature of spacetime then how are we to measure, understand and explain this curvature? Of course humans have experienced a similar problem before: the surface of the Earth is a two-dimensional space and there are several means of working out that the Earth is, to a good approximation, a spherical ball. 1 1 ^(1){ }^{1}1
Carl Friedrich Gauss made some of the most significant progress in finding that there was a way to convert measurements of distances, made by an observer trapped on a two-dimensional surface, into an objective description of the curvature of that space. However, Bernhard Riemann 2 2 ^(2){ }^{2}2 was engaged in an even more ambitious programme to evaluate the curvature of higher dimensional spaces. He solved the problem in 1854 and introduced a mathematical description that would be built up by Christoffel, Ricci, Levi-Civita and others. It is this description that was incorporated into physics by Einstein. Riemann's description of curvature is encoded into a tensor R R R\boldsymbol{R}R. We shall see that a non-zero R R R\boldsymbol{R}R is the sure-fire way of telling mathematically that a spacetime is truly curved. 3 3 ^(3){ }^{3}3

11.1 What is curvature?

Euclid's fifth postulate, known as the parallel postulate, may be paraphrased as saying that parallel lines always remain parallel. Euclid did not prove this, which is why it remained a postulate. It turns out only to be the case in flat space.
The surface of a cylinder appears curved, but this curvature is extrinsic: we can recreate (or wrap) the curved surface of the cylinder using a flat piece of paper with two of its opposite edges identified. We can therefore unravel the cylinder, turning it into a flat plane. Parallel lines on a cylindrical surface never meet, therefore, as they are equivalent to parallel lines on the flat plane. The sphere is different, its curvature is intrinsic: it is impossible to wrap the sphere in a flat piece of paper in the way that we did the cylinder. Parallel lines on the surface of a sphere don't remain parallel; they eventually meet.
Curved surfaces can be classified in terms of the paths of such parallel lines. On a flat surface parallel lines remain parallel. On a surface with positive curvature initially parallel lines converge; on a surface with negative curvature they diverge (Fig. 11.1). An example of a surface with positive curvature is the surface of a sphere; surfaces with negative curvature can be thought of as being saddle-like.
There are two simple ways of telling whether curvature is intrinsic or extrinsic. The first, motivated by the discussion so far, is to set two free, spatially separated particles moving parallel. If the space is intrinsically curved, the particles' paths will deviate from parallelism. This effect, shown in Fig. 11.1 is known as geodesic deviation.
The second method is to parallel transport a vector in a closed loop. If the space has intrinsic curvature, then the vector will rotate. If the space has extrinsic curvature, there will be no change in direction. The rotation is most easily seen by examining parallel transport on the surface of a sphere as shown in Fig. 11.2.
In summary, to measure the amount of curvature present we choose one of the two methods: (i) identify a fiducial geodesic and compare it to another geodesic whose tangent vector was originally parallel. The extent of the deviation tells us the amount by which the space is curved. (ii) Take a vector and parallel transport it around a loop in spacetime. The amount by which the vector changes its direction provides a measure of curvature. As we shall show, these two methods both provide the same measure of curvature: the components of the Riemann curvature tensor R R R\boldsymbol{R}R.
The ultimate source of curvature is the metric field. We saw in the previous chapters how the connection coefficients were derived from the first derivatives of the components of the metric (with the rule of thumb expression that g ) ( Γ ¯ ) g ) ( Γ ¯ ) del g)rarr( bar(Gamma))\partial g) \rightarrow(\bar{\Gamma})g)(Γ¯) ). The Riemann curvature tensor is built from the first and second derivatives of the metric and so we have the rule of thumb
(11.1) ( g ) + 2 g ) ( R ¯ ) (11.1) ( g ¯ ) + 2 g ( R ¯ ) {:(11.1){:( bar(del g))+del^(del^(2)g))rarr( bar(R)):}\begin{equation*} \left.(\overline{\partial g})+\partial^{\partial^{2} g}\right) \rightarrow(\bar{R}) \tag{11.1} \end{equation*}(11.1)(g)+2g)(R¯)
This leads us to expect that the components of this tensor can also be written in terms of the connection coefficients and their derivatives (i.e. ( Γ ¯ ) + ( Γ ) ( R ) ( Γ ¯ ) + ( Γ _ ) ( R _ ) ( bar(Gamma))+(del Gamma_)rarr(R_)(\bar{\Gamma})+(\underline{\partial \Gamma}) \rightarrow(\underline{R})(Γ¯)+(Γ)(R), which, as we shall see, turns out to be the case.

11.2 Tidal forces

The simplest gravitating system might be expected to have a static metric, completely independent of time. This might represent the geometry around a star. Such a geometry would allow us to identify a static frame of reference (i.e. the rest frame of the star). We could then perform experiments in which we placed a single particle at rest in this frame and watched to see if it started accelerating. This would allow us to identify a gravitational field. However, this is not a good description of our Universe, where the principles of relativity teach us that it is not generally possible to identify a static frame against which to identify accelera-
Euclid's postulates are:
I: A straight line segment can be drawn joining any two points.
II: Any straight line segment can be extended indefinitely in a straight line. III: Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as centre. IV: All right angles are congruent. V: If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect the two lines inevitably must intersect
each other on that side if extended far each other on that side if extended far
enough. This postulate is equivalent to the parallel postulate.
Fig. 11.1 Geodesic deviation. In a curved spacetime, particles fall along geodesics that are parallel at some point in spacetime. Extending the point in spacetime. Extending the paths into the future and past, the
world lines of the particles meet if the space is positively curved.
Fig. 11.2 Transporting a vector around a closed loop on the surface of a around a closed loop on the surface of a
sphere. The vector rotates as a result of the intrinsic curvature of the spherical surface.
4 4 ^(4){ }^{4}4 We prove this statement about the volume staying constant at the end of this section.
Fig. 11.3 Tidal forces. (a) The forces on a cloud of particles surrounding an observer that are falling towards the Earth, shown in the Earth's rest frame. (b) The same forces in the rest frame of the observer.
Fig. 11.4 Particles falling in a Newtonian potential, with separation vector s s vec(s)\vec{s}s.
tion. We saw that taking relativity into account leads to the equivalence principle, telling us that from the motion of a single particle we cannot tell the difference between acceleration due to a choice of coordinates and acceleration due to spacetime curvature. The best we can do, is to examine the relative motion of two particles.
Let's briefly return to the world of Newtonian gravity and discuss geodesic deviation in this limit. Consider a spherical cluster of test particles falling towards the earth. Each particle moves on a straight line through the Earth's centre, but those that are closer fall faster than those further away, owing to the greater force on the closer particles. As a result of the motion, the sphere won't remain spherical, but will be distorted into an ellipse of the same volume. 4 4 ^(4){ }^{4}4 In the rest frame of an observer falling with the particles, the forces that cause the change to the shape of bodies in a gravitational field are known as tidal forces. They are illustrated in Fig. 11.3.

Example 11.1

Consider a cloud of dust particles that are initially spherical, that are released in the vicinity of a gravitating mass. Newton's second law gives us the equation of motion for each particle in a gravitational potential Φ ( x ) = G M / | x | Φ ( x ) = G M / | x | Phi( vec(x))=-GM//| vec(x)|\Phi(\vec{x})=-G M /|\vec{x}|Φ(x)=GM/|x| (temporarily relaxing our rules on balanced up and down tensor components)
(11.2) m d 2 x i d t 2 = m Φ ( x ) x i (11.2) m d 2 x i d t 2 = m Φ ( x ) x i {:(11.2)m(d^(2)x^(i))/((d)t^(2))=-m(del Phi(( vec(x))))/(delx^(i)):}\begin{equation*} m \frac{\mathrm{~d}^{2} x^{i}}{\mathrm{~d} t^{2}}=-m \frac{\partial \Phi(\vec{x})}{\partial x^{i}} \tag{11.2} \end{equation*}(11.2)m d2xi dt2=mΦ(x)xi
We imagine the trajectories of two particles in the cloud, shown in Fig. 11.4, with a vector s ( t ) = y ( t ) x ( t ) s ( t ) = y ( t ) x ( t ) vec(s)(t)= vec(y)(t)- vec(x)(t)\vec{s}(t)=\vec{y}(t)-\vec{x}(t)s(t)=y(t)x(t) stretching between them. (The connecting vector is assumed small compared to the distance from the gravitating mass in what follows.) We can take the difference in their equations of motion
( d 2 y i d t 2 d 2 x i d t 2 ) = Φ ( y ) x i + Φ ( x ) x i = Φ ( x + s ) x i + Φ ( x ) x i . (11.3) 2 Φ ( x ) x i x j s j . d 2 y i d t 2 d 2 x i d t 2 = Φ ( y ) x i + Φ ( x ) x i = Φ ( x + s ) x i + Φ ( x ) x i . (11.3) 2 Φ ( x ) x i x j s j . {:[((d^(2)y^(i))/((d)t^(2))-(d^(2)x^(i))/((d)t^(2)))=-(del Phi(( vec(y))))/(delx^(i))+(del Phi(( vec(x))))/(delx^(i))],[=-(del Phi(( vec(x))+( vec(s))))/(delx^(i))+(del Phi(( vec(x))))/(delx^(i)).],[(11.3)~~-(del^(2)Phi(( vec(x))))/(delx^(i)delx^(j))s^(j).]:}\begin{align*} \left(\frac{\mathrm{d}^{2} y^{i}}{\mathrm{~d} t^{2}}-\frac{\mathrm{d}^{2} x^{i}}{\mathrm{~d} t^{2}}\right) & =-\frac{\partial \Phi(\vec{y})}{\partial x^{i}}+\frac{\partial \Phi(\vec{x})}{\partial x^{i}} \\ & =-\frac{\partial \Phi(\vec{x}+\vec{s})}{\partial x^{i}}+\frac{\partial \Phi(\vec{x})}{\partial x^{i}} . \\ & \approx-\frac{\partial^{2} \Phi(\vec{x})}{\partial x^{i} \partial x^{j}} s^{j} . \tag{11.3} \end{align*}(d2yi dt2d2xi dt2)=Φ(y)xi+Φ(x)xi=Φ(x+s)xi+Φ(x)xi.(11.3)2Φ(x)xixjsj.
Define a tensor with components R i j = R i j = 2 Φ ( x ) x i x j R i j = R i j = 2 Φ ( x ) x i x j R^(i)_(j)=R_(ij)=(del^(2)Phi(( vec(x))))/(delx^(i)delx^(j))R^{i}{ }_{j}=R_{i j}=\frac{\partial^{2} \Phi(\vec{x})}{\partial x^{i} \partial x^{j}}Rij=Rij=2Φ(x)xixj, leading to the equation of motion of the components of the deviation vector
(11.4) d 2 s i d t 2 = R j i s j (11.4) d 2 s i d t 2 = R j i s j {:(11.4)(d^(2)s^(i))/((d)t^(2))=-R_(j)^(i)s^(j):}\begin{equation*} \frac{\mathrm{d}^{2} s^{i}}{\mathrm{~d} t^{2}}=-R_{j}^{i} s^{j} \tag{11.4} \end{equation*}(11.4)d2si dt2=Rjisj
Let's examine the case where the particle starts with x = ( 0 , 0 , z ) x = ( 0 , 0 , z ) vec(x)=(0,0,z)\vec{x}=(0,0, z)x=(0,0,z), and falls towards a planet, whose centre is the origin. The tensor R i j R i j R^(i)_(j)R^{i}{ }_{j}Rij may be evaluated, giving
(11.5) R j i = G M z 3 ( 1 0 0 0 1 0 0 0 2 ) (11.5) R j i = G M z 3 1 0 0 0 1 0 0 0 2 {:(11.5)R_(j)^(i)=(GM)/(z^(3))([1,0,0],[0,1,0],[0,0,-2]):}R_{j}^{i}=\frac{G M}{z^{3}}\left(\begin{array}{ccc} 1 & 0 & 0 \tag{11.5}\\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{array}\right)(11.5)Rji=GMz3(100010002)
The cloud becomes elongated along z z zzz, because d 2 s z d t 2 = 2 G M r 3 z d 2 s z d t 2 = 2 G M r 3 z (d^(2)s^(z))/((d)t^(2))=2(GM)/(r^(3))z\frac{\mathrm{d}^{2} s^{z}}{\mathrm{~d} t^{2}}=2 \frac{G M}{r^{3}} zd2sz dt2=2GMr3z. The cloud is compressed along x x xxx and y y yyy, this is because d 2 s i d t 2 = G M r 3 s i d 2 s i d t 2 = G M r 3 s i (d^(2)s^(i))/((d)t^(2))=-(GM)/(r^(3))*s^(i)\frac{\mathrm{d}^{2} s^{i}}{\mathrm{~d} t^{2}}=-\frac{G M}{r^{3}} \cdot s^{i}d2si dt2=GMr3si, for i = x i = x i=xi=xi=x and y y yyy.
After this discussion of Newtonian mechanics, we turn to relativistic gravitation. We use the same strategy, but this time using the Riemann tensor R R R\boldsymbol{R}R. The technique is, once again, to assess the geodesic deviation. We identify some freely falling particle on a fiducial geodesic and note its 4 -velocity u u u\boldsymbol{u}u. We compare the geodesic of a different freely falling particle on a different geodesic, whose velocity was originally parallel to the fiducial one. The separation of the two geodesics is ξ ξ xi\boldsymbol{\xi}ξ. If there's curvature present then ξ ξ xi\boldsymbol{\xi}ξ will accelerate. The equation of motion for the separation vector is 5 5 ^(5){ }^{5}5
(11.6) D 2 ξ d τ 2 + R ( , u , ξ , u ) = 0 (11.6) D 2 ξ d τ 2 + R ( , u , ξ , u ) = 0 {:(11.6)(D^(2)xi)/((d)tau^(2))+R(","u","xi","u)=0:}\begin{equation*} \frac{\mathrm{D}^{2} \boldsymbol{\xi}}{\mathrm{~d} \tau^{2}}+\boldsymbol{R}(, \boldsymbol{u}, \boldsymbol{\xi}, \boldsymbol{u})=0 \tag{11.6} \end{equation*}(11.6)D2ξ dτ2+R(,u,ξ,u)=0
The (1,3) tensor R ( , , R ( , , R(,,\boldsymbol{R}(,,R(,,, ) i s t h e k e y o b j e c t f o r a s s e s s i n g c u r v a t u r e . ) i s t h e k e y o b j e c t f o r a s s e s s i n g c u r v a t u r e . )isthekeyobjectforassessingcurvature.) is the key object for assessing curvature.)isthekeyobjectforassessingcurvature. Filling the latter three slots with vectors as shown in eqn 11.6 gives rise to a ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) object (i.e. a vector) that measures the relative acceleration of the geodesics. If R = 0 R = 0 R=0\boldsymbol{R}=0R=0 (that is, its components vanish) then space is flat. If the components aren't zero then we are able to say that the space is curved. 6 6 ^(6){ }^{6}6
We shall return to the use of geodesic deviation at the end of the chapter. As a measure of curvature it is conceptually clear, but computations rely on some mathematical tricks we haven't yet encountered. In the next section we make use of the other approach, parallel transport around a loop, in order to obtain an explicit form of the tensor R R R\boldsymbol{R}R.

Example 11.2

Proof that the volume of the cluster of particles is constant as it deforms. Recall that we can write Coulomb's law for the electrostatic force between charges and justify Gauss' law for the divergence of the E E vec(E)\vec{E}E-field from charge density ρ c ρ c rho_(c)\rho_{\mathrm{c}}ρc, which reads E = ρ c / ε 0 E = ρ c / ε 0 vec(grad)* vec(E)=rho_(c)//epsi_(0)\vec{\nabla} \cdot \vec{E}=\rho_{\mathrm{c}} / \varepsilon_{0}E=ρc/ε0. In the same way, we can write a differential equation relating the gravitational field 7 g 7 g ^(7) vec(g){ }^{7} \vec{g}7g to a density of matter ρ ρ rho\rhoρ, which reads g = 4 π G ρ g = 4 π G ρ vec(grad)* vec(g)=-4pi G rho\vec{\nabla} \cdot \vec{g}=-4 \pi G \rhog=4πGρ. In integral form, this is written as
(11.7) g d S = 4 π G M , (11.7) g d S = 4 π G M , {:(11.7)int vec(g)*d vec(S)=-4pi GM",":}\begin{equation*} \int \vec{g} \cdot \mathrm{~d} \vec{S}=-4 \pi G M, \tag{11.7} \end{equation*}(11.7)g dS=4πGM,
where M M MMM is the total mass in a volume bounded by the closed surface with area vector S S vec(S)\vec{S}S. In words this expression says that the strength of the gravitational field is given by the mass inside the surface S S vec(S)\vec{S}S. If, instead of g g vec(g)\vec{g}g we use the potential Φ Φ Phi\PhiΦ, via g = Φ g = Φ vec(g)=- vec(grad)Phi\vec{g}=-\vec{\nabla} \Phig=Φ, then we obtain Poisson's equation 8 8 ^(8){ }^{8}8 for gravitational fields,
(11.8) 2 Φ = 4 π G ρ . (11.8) 2 Φ = 4 π G ρ . {:(11.8) vec(grad)^(2)Phi=4pi G rho.:}\begin{equation*} \vec{\nabla}^{2} \Phi=4 \pi G \rho . \tag{11.8} \end{equation*}(11.8)2Φ=4πGρ.
We can use this expression to justify the claim that a sphere of test particles falling towards the earth does not change its volume. Note first that 2 Φ 2 Φ grad^(2)Phi\nabla^{2} \Phi2Φ is the sum of eigenvalues of the matrix 2 Φ x i x 3 2 Φ x i x 3 (del^(2)Phi)/(delx^(i)delx^(3))\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{3}}2Φxix3 that we considered above. We can then use the intervals s i s i s^(i)s^{i}si from the previous example to build a small sphere of particles. We then have, as a result of our expression m s i = 2 Φ x i x j s j m s i = 2 Φ x i x j s j ms^(i)=-(del^(2)Phi)/(delx^(i)delx^(j))s^(j)m s^{i}=-\frac{\partial^{2} \Phi}{\partial x^{i} \partial x^{j}} s^{j}msi=2Φxixjsj, that the second time derivative of the volume of the sphere of particles is determined by the density ρ ρ rho\rhoρ of matter inside the sphere. Therefore, a sphere of free particles falling towards the earth forms an ellipsoid of the same volume as the original sphere, as there is (always) no gravitating mass inside the sphere. (If, on the other hand, we built a large sphere around the earth, we would expect the sphere to reduce its volume as the particles fell towards the earth.)
5 5 ^(5){ }^{5}5 This equation is discussed in detail in Chapter 35 where it is justified more thoroughly and compared against the parallel transport method discussed in the chapter.
6 6 ^(6){ }^{6}6 The various methods of measuring curvature are discused in detail in curvature are discussed in detail in Chapter 30 . As a rough measure, we
say that the amount of curvature induced by the non-zero Riemann tensor duced by the non-zero Riemann tensor
is, in the orthonormal frame, of order is, in the orth
( 1 / R μ ^ ν ^ α ^ β ^ 1 2 ) 1 2 1 / R μ ^ ν ^ α ^ β ^ 1 2 1 2 (1//R^( hat(mu))_( hat(nu) hat(alpha) hat(beta))^((1)/(2)))^((1)/(2))\left(1 / R^{\hat{\mu}}{ }_{\hat{\nu} \hat{\alpha} \hat{\beta}}^{\frac{1}{2}}\right)^{\frac{1}{2}}(1/Rμ^ν^α^β^12)12.
7 7 ^(7){ }^{7}7 Defined by F = m g F = m g vec(F)=m vec(g)\vec{F}=m \vec{g}F=mg, where F F vec(F)\vec{F}F is the gravitation force on a particle with mass m m mmm.

11.3 Riemann curvature

Although geodesic deviation provides a physical and intuitive means of working out if space is curved, there is a mathematically simpler 9 9 ^(9){ }^{9}9 method of determining the form of the Riemann tensor, involving the parallel transport of a vector. Recall that parallel transport is a method to work out how a vector changes due to the change in the underlying coordinate system. This isn't curvature necessarily, we may just be using a hopelessly complicated coordinate system. In order to work out
Fig. 11.5 Transporting a vector around an infinitesimal closed loop. if a spacetime is truly curved, we recall the fate of a vector parallel transported on a spherical surface from Fig. 11.2. We can see from this figure the effect of parallel transportation around a closed loop: the vector rotates. This is a smoking gun, telling us that the surface is truly curved. If it were not, the vector would point in the same direction after parallel transport. In order to assess the curvature of a spacetime we shall follow the same procedure and move a 4 -vector in an infinitesimal loop to see if its direction changes. If it does, we will be able to identify the intrinsic curvature of the spacetime.
So let's be specific and take a vector V V V\boldsymbol{V}V and move it round a closed, infinitesimal parallelogram formed by a vectors δ a e 1 δ a e 1 delta ae_(1)\delta a \boldsymbol{e}_{1}δae1 and δ b e 2 δ b e 2 delta be_(2)\delta b \boldsymbol{e}_{2}δbe2, where δ a δ a delta a\delta aδa and δ b δ b delta b\delta bδb are (small) constants (see Fig. 11.5). The idea is that we track the change in a component of V V V\boldsymbol{V}V as follows:
δ V α = ( Change in V α transported along δ a e 1 , then δ b e 2 , then δ a e 1 , then δ b e 2 ) (11.9) = ( Measure of the curvature of spacetime ) μ V μ δ a δ b . δ V α = (  Change in  V α  transported along  δ a e 1 ,  then  δ b e 2 ,  then  δ a e 1 ,  then  δ b e 2 ) (11.9) = (  Measure of the curvature   of spacetime  ) μ V μ δ a δ b . {:[deltaV^(alpha)=((" Change in "V^(alpha)" transported along "delta ae_(1),)/(" then "delta be_(2)," then "-delta ae_(1)," then "-delta be_(2)))],[(11.9)=((" Measure of the curvature ")/(" of spacetime "))_(mu)V^(mu)delta a delta b.]:}\begin{align*} \delta V^{\alpha} & =\binom{\text { Change in } V^{\alpha} \text { transported along } \delta a \boldsymbol{e}_{1},}{\text { then } \delta b \boldsymbol{e}_{2}, \text { then }-\delta a \boldsymbol{e}_{1}, \text { then }-\delta b \boldsymbol{e}_{2}} \\ & =\binom{\text { Measure of the curvature }}{\text { of spacetime }}_{\mu} V^{\mu} \delta a \delta b . \tag{11.9} \end{align*}δVα=( Change in Vα transported along δae1, then δbe2, then δae1, then δbe2)(11.9)=( Measure of the curvature  of spacetime )μVμδaδb.
This measure of curvature is, once again, supplied by the components of the Riemann tensor R ( , , R ( , , R(,,\boldsymbol{R}(,,R(,,, ) . H o w d o e s t h i s w o r k ? I n t h e p r e c e d ) . H o w d o e s t h i s w o r k ? I n t h e p r e c e d ).Howdoesthiswork?Inthepreced-) . How does this work? In the preced-).Howdoesthiswork?Inthepreced ing equation, we have made the deceptively simple, but mathematically significant, claim that the right-hand side of the equation is linear in the sides of the parallelogram δ a δ a delta a\delta aδa and δ b δ b delta b\delta bδb. It is this that enables us to identify the part in brackets as the components of a tensor (i.e. a linear slot-machine object). To show that this is the case, we will dive ahead and use this method to derive an expression for the curvature tensor.

Example 11.3

We start with the vector at position A A AAA, with coordinates ( a , b ) ( a , b ) (a,b)(a, b)(a,b) and parallel transport it along e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 to position B B BBB. Parallel transport along e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 implies e 1 V = 0 e 1 V = 0 grad_(e_(1))V=0\boldsymbol{\nabla}_{\boldsymbol{e}_{1}} \boldsymbol{V}=0e1V=0, so we must have
(11.10) V α x 1 = Γ 1 β α V β (11.10) V α x 1 = Γ 1 β α V β {:(11.10)(delV^(alpha))/(delx^(1))=-Gamma_(1beta)^(alpha)V^(beta):}\begin{equation*} \frac{\partial V^{\alpha}}{\partial x^{1}}=-\Gamma_{1 \beta}^{\alpha} V^{\beta} \tag{11.10} \end{equation*}(11.10)Vαx1=Γ1βαVβ
At position B B BBB, the vector has components
V α ( B ) = V α ( A ) + A B d x 1 V α x 1 (11.11) = V α ( A ) x 2 = b d x 1 Γ α 1 β V β . V α ( B ) = V α ( A ) + A B d x 1 V α x 1 (11.11) = V α ( A ) x 2 = b d x 1 Γ α 1 β V β . {:[V^(alpha)(B)=V^(alpha)(A)+int_(A)^(B)dx^(1)(delV^(alpha))/(delx^(1))],[(11.11)=V^(alpha)(A)-int_(x^(2)=b)dx^(1)Gamma^(alpha)_(1beta)V^(beta).]:}\begin{align*} V^{\alpha}(B) & =V^{\alpha}(A)+\int_{A}^{B} \mathrm{~d} x^{1} \frac{\partial V^{\alpha}}{\partial x^{1}} \\ & =V^{\alpha}(A)-\int_{x^{2}=b} \mathrm{~d} x^{1} \Gamma^{\alpha}{ }_{1 \beta} V^{\beta} . \tag{11.11} \end{align*}Vα(B)=Vα(A)+AB dx1Vαx1(11.11)=Vα(A)x2=b dx1Γα1βVβ.
We repeat, transporting to points C C CCC and D D DDD and back to A 10 A 10 A^(10)A^{10}A10
V α ( C ) = V α ( B ) x 1 = a + δ a d x 2 Γ α 2 β V β , V α ( D ) = V α ( C ) + x 2 = b + δ b d x 1 Γ α 1 β V β , (11.12) V α ( A final ) = V α ( D ) + x 1 = a d x 2 Γ α 2 β V β . V α ( C ) = V α ( B ) x 1 = a + δ a d x 2 Γ α 2 β V β , V α ( D ) = V α ( C ) + x 2 = b + δ b d x 1 Γ α 1 β V β , (11.12) V α A final  = V α ( D ) + x 1 = a d x 2 Γ α 2 β V β . {:[V^(alpha)(C)=V^(alpha)(B)-int_(x^(1)=a+delta a)dx^(2)Gamma^(alpha)_(2beta)V^(beta)","],[V^(alpha)(D)=V^(alpha)(C)+int_(x^(2)=b+delta b)dx^(1)Gamma^(alpha)_(1beta)V^(beta)","],[(11.12)V^(alpha)(A_("final "))=V^(alpha)(D)+int_(x^(1)=a)dx^(2)Gamma^(alpha)_(2beta)V^(beta).]:}\begin{align*} V^{\alpha}(C) & =V^{\alpha}(B)-\int_{x^{1}=a+\delta a} \mathrm{~d} x^{2} \Gamma^{\alpha}{ }_{2 \beta} V^{\beta}, \\ V^{\alpha}(D) & =V^{\alpha}(C)+\int_{x^{2}=b+\delta b} \mathrm{~d} x^{1} \Gamma^{\alpha}{ }_{1 \beta} V^{\beta}, \\ V^{\alpha}\left(A_{\text {final }}\right) & =V^{\alpha}(D)+\int_{x^{1}=a} \mathrm{~d} x^{2} \Gamma^{\alpha}{ }_{2 \beta} V^{\beta} . \tag{11.12} \end{align*}Vα(C)=Vα(B)x1=a+δa dx2Γα2βVβ,Vα(D)=Vα(C)+x2=b+δb dx1Γα1βVβ,(11.12)Vα(Afinal )=Vα(D)+x1=a dx2Γα2βVβ.
The net change δ V α δ V α deltaV^(alpha)\delta V^{\alpha}δVα going round the loop is, therefore,
δ V α = V α ( A final ) V α ( A initial ) = x 1 = a d x 2 Γ α 2 β V β x 1 = a + δ a d x 2 Γ α 2 β V β (11.13) + x 2 = b + δ b d x 1 Γ α 1 β V β x 2 = b d x 1 Γ α 1 β V β δ V α = V α A final  V α A initial  = x 1 = a d x 2 Γ α 2 β V β x 1 = a + δ a d x 2 Γ α 2 β V β (11.13) + x 2 = b + δ b d x 1 Γ α 1 β V β x 2 = b d x 1 Γ α 1 β V β {:[deltaV^(alpha)=V^(alpha)(A_("final "))-V^(alpha)(A_("initial "))],[=int_(x^(1)=a)dx^(2)Gamma^(alpha)_(2beta)V^(beta)-int_(x^(1)=a+delta a)dx^(2)Gamma^(alpha)_(2beta)V^(beta)],[(11.13)+int_(x^(2)=b+delta b)dx^(1)Gamma^(alpha)_(1beta)V^(beta)-int_(x^(2)=b)dx^(1)Gamma^(alpha)_(1beta)V^(beta)]:}\begin{align*} \delta V^{\alpha}= & V^{\alpha}\left(A_{\text {final }}\right)-V^{\alpha}\left(A_{\text {initial }}\right) \\ = & \int_{x^{1}=a} \mathrm{~d} x^{2} \Gamma^{\alpha}{ }_{2 \beta} V^{\beta}-\int_{x^{1}=a+\delta a} \mathrm{~d} x^{2} \Gamma^{\alpha}{ }_{2 \beta} V^{\beta} \\ & +\int_{x^{2}=b+\delta b} \mathrm{~d} x^{1} \Gamma^{\alpha}{ }_{1 \beta} V^{\beta}-\int_{x^{2}=b} \mathrm{~d} x^{1} \Gamma^{\alpha}{ }_{1 \beta} V^{\beta} \tag{11.13} \end{align*}δVα=Vα(Afinal )Vα(Ainitial )=x1=a dx2Γα2βVβx1=a+δa dx2Γα2βVβ(11.13)+x2=b+δb dx1Γα1βVβx2=b dx1Γα1βVβ
Collecting terms and expanding to lowest order, we obtain
δ V α b b + δ b d x 2 δ a x 1 ( Γ α 2 β V β ) + a a + δ a d x 1 δ b x 2 ( Γ α 1 β V β ) (11.14) δ a δ b [ x 1 ( Γ α 2 β V β ) + x 2 ( Γ α 1 β V β ) ] . δ V α b b + δ b d x 2 δ a x 1 Γ α 2 β V β + a a + δ a d x 1 δ b x 2 Γ α 1 β V β (11.14) δ a δ b x 1 Γ α 2 β V β + x 2 Γ α 1 β V β . {:[deltaV^(alpha)~~-int_(b)^(b+delta b)dx^(2)delta a(del)/(delx^(1))(Gamma^(alpha)_(2beta)V^(beta))+int_(a)^(a+delta a)dx^(1)delta b(del)/(delx^(2))(Gamma^(alpha)_(1beta)V^(beta))],[(11.14)~~delta a delta b[-(del)/(delx^(1))(Gamma^(alpha)_(2beta)V^(beta))+(del)/(delx^(2))(Gamma^(alpha)_(1beta)V^(beta))].]:}\begin{align*} \delta V^{\alpha} & \approx-\int_{b}^{b+\delta b} \mathrm{~d} x^{2} \delta a \frac{\partial}{\partial x^{1}}\left(\Gamma^{\alpha}{ }_{2 \beta} V^{\beta}\right)+\int_{a}^{a+\delta a} \mathrm{~d} x^{1} \delta b \frac{\partial}{\partial x^{2}}\left(\Gamma^{\alpha}{ }_{1 \beta} V^{\beta}\right) \\ & \approx \delta a \delta b\left[-\frac{\partial}{\partial x^{1}}\left(\Gamma^{\alpha}{ }_{2 \beta} V^{\beta}\right)+\frac{\partial}{\partial x^{2}}\left(\Gamma^{\alpha}{ }_{1 \beta} V^{\beta}\right)\right] . \tag{11.14} \end{align*}δVαbb+δb dx2δax1(Γα2βVβ)+aa+δa dx1δbx2(Γα1βVβ)(11.14)δaδb[x1(Γα2βVβ)+x2(Γα1βVβ)].
Finally, replacing the derivatives of V α V α V^(alpha)V^{\alpha}Vα using eqn 11.10, we obtain
(11.15) δ V α δ a δ b [ Γ 1 β α x 2 Γ 2 β α x 1 + Γ 2 σ α Γ 1 β σ Γ 1 σ α Γ 2 β σ ] V β (11.15) δ V α δ a δ b Γ 1 β α x 2 Γ 2 β α x 1 + Γ 2 σ α Γ 1 β σ Γ 1 σ α Γ 2 β σ V β {:(11.15)deltaV^(alpha)~~delta a delta b[(delGamma_(1beta)^(alpha))/(delx^(2))-(delGamma_(2beta)^(alpha))/(delx^(1))+Gamma_(2sigma)^(alpha)Gamma_(1beta)^(sigma)-Gamma_(1sigma)^(alpha)Gamma_(2beta)^(sigma)]V^(beta):}\begin{equation*} \delta V^{\alpha} \approx \delta a \delta b\left[\frac{\partial \Gamma_{1 \beta}^{\alpha}}{\partial x^{2}}-\frac{\partial \Gamma_{2 \beta}^{\alpha}}{\partial x^{1}}+\Gamma_{2 \sigma}^{\alpha} \Gamma_{1 \beta}^{\sigma}-\Gamma_{1 \sigma}^{\alpha} \Gamma_{2 \beta}^{\sigma}\right] V^{\beta} \tag{11.15} \end{equation*}(11.15)δVαδaδb[Γ1βαx2Γ2βαx1+Γ2σαΓ1βσΓ1σαΓ2βσ]Vβ
The change in the vector V V V\boldsymbol{V}V can be summarized as
(11.16) δ V α = ( Change in V α transported along δ a e ν , then δ b e μ , then δ a e ν , then δ b e μ ) = [ Γ ν β α x μ Γ α μ β x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ ] V β δ a δ b . (11.16) δ V α = (  Change in  V α  transported along  δ a e ν ,  then  δ b e μ ,  then  δ a e ν ,  then  δ b e μ ) = Γ ν β α x μ Γ α μ β x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ V β δ a δ b . {:[(11.16)deltaV^(alpha)=((" Change in "V^(alpha)" transported along "delta ae_(nu),)/(" then "delta be_(mu)," then "-delta ae_(nu)," then "-delta be_(mu)))],[=[(delGamma_(nu beta)^(alpha))/(delx^(mu))-(delGamma^(alpha)_(mu beta))/(delx^(nu))+Gamma_(mu sigma)^(alpha)Gamma_(nu beta)^(sigma)-Gamma_(nu sigma)^(alpha)Gamma_(mu beta)^(sigma)]V^(beta)delta a delta b.]:}\begin{align*} \delta V^{\alpha} & =\binom{\text { Change in } V^{\alpha} \text { transported along } \delta a \boldsymbol{e}_{\nu},}{\text { then } \delta b \boldsymbol{e}_{\mu}, \text { then }-\delta a \boldsymbol{e}_{\nu}, \text { then }-\delta b \boldsymbol{e}_{\mu}} \tag{11.16}\\ & =\left[\frac{\partial \Gamma_{\nu \beta}^{\alpha}}{\partial x^{\mu}}-\frac{\partial \Gamma^{\alpha}{ }_{\mu \beta}}{\partial x^{\nu}}+\Gamma_{\mu \sigma}^{\alpha} \Gamma_{\nu \beta}^{\sigma}-\Gamma_{\nu \sigma}^{\alpha} \Gamma_{\mu \beta}^{\sigma}\right] V^{\beta} \delta a \delta b . \end{align*}(11.16)δVα=( Change in Vα transported along δaeν, then δbeμ, then δaeν, then δbeμ)=[ΓνβαxμΓαμβxν+ΓμσαΓνβσΓνσαΓμβσ]Vβδaδb.
As claimed, the part in square brackets depends linearly on the sides of the parallelogram δ a δ a delta a\delta aδa and δ b δ b delta b\delta bδb. This enables us to define the Riemann curvature tensor as the (1,3) tensor R ( , R , R(,:}\boldsymbol{R}\left(,\right.R(,, , with components 11 11 ^(11){ }^{11}11
(11.19) R β μ ν α = [ Γ ν β α x μ Γ μ β α x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ ] (11.19) R β μ ν α = Γ ν β α x μ Γ μ β α x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ {:(11.19)R_(beta mu nu)^(alpha)=[(delGamma_(nu beta)^(alpha))/(delx^(mu))-(delGamma_(mu beta)^(alpha))/(delx^(nu))+Gamma_(mu sigma)^(alpha)Gamma_(nu beta)^(sigma)-Gamma_(nu sigma)^(alpha)Gamma_(mu beta)^(sigma)]:}\begin{equation*} R_{\beta \mu \nu}^{\alpha}=\left[\frac{\partial \Gamma_{\nu \beta}^{\alpha}}{\partial x^{\mu}}-\frac{\partial \Gamma_{\mu \beta}^{\alpha}}{\partial x^{\nu}}+\Gamma_{\mu \sigma}^{\alpha} \Gamma_{\nu \beta}^{\sigma}-\Gamma_{\nu \sigma}^{\alpha} \Gamma_{\mu \beta}^{\sigma}\right] \tag{11.19} \end{equation*}(11.19)Rβμνα=[ΓνβαxμΓμβαxν+ΓμσαΓνβσΓνσαΓμβσ]

Example 11.4

Recall for the weak gravitational field metric d s 2 = [ 1 + 2 Φ ( x i ) ] d t 2 + ( d x 2 + d y 2 + d s 2 = 1 + 2 Φ x i d t 2 + d x 2 + d y 2 + ds^(2)=-[1+2Phi(x^(i))]dt^(2)+(dx^(2)+dy^(2)+:}\mathrm{d} s^{2}=-\left[1+2 \Phi\left(x^{i}\right)\right] \mathrm{d} t^{2}+\left(\mathrm{d} x^{2}+\mathrm{d} y^{2}+\right.ds2=[1+2Φ(xi)]dt2+(dx2+dy2+ d z 2 d z 2 dz^(2)\mathrm{d} z^{2}dz2 ) we had non-zero connections
(11.20) Γ i t t = Φ x i . (11.20) Γ i t t = Φ x i . {:(11.20)Gamma^(i)_(tt)=(del Phi)/(delx^(i)).:}\begin{equation*} \Gamma^{i}{ }_{t t}=\frac{\partial \Phi}{\partial x^{i}} . \tag{11.20} \end{equation*}(11.20)Γitt=Φxi.
Plugging in, we find that the only non-zero components of R R R\boldsymbol{R}R are given by 12 12 ^(12){ }^{12}12
(11.21) R t j t i = Γ t t i x j = 2 Φ x j x i . (11.21) R t j t i = Γ t t i x j = 2 Φ x j x i . {:(11.21)R_(tjt)^(i)=(delGamma_(tt)^(i))/(delx^(j))=(del^(2)Phi)/(delx^(j)delx^(i)).:}\begin{equation*} R_{t j t}^{i}=\frac{\partial \Gamma_{t t}^{i}}{\partial x^{j}}=\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{i}} . \tag{11.21} \end{equation*}(11.21)Rtjti=Γttixj=2Φxjxi.
These are identical to the components that we identified earlier in the chapter in eqn 11.4.
10 10 ^(10){ }^{10}10 The notation here records the constant coordinate, rather than the limits of the integral.
11 11 ^(11){ }^{11}11 Equation 11.19 has the dubious distinction of being one of the more lengthy equations in this subject that it is useful to remember. A helpful aidemémoire that is often employed is to use the matrix form of the equations to say
R μ ν = μ Γ ν ν Γ μ + [ Γ μ , Γ ν ] , 11.17 ) R μ ν = μ Γ ν ν Γ μ + Γ μ , Γ ν , 11.17 ) {:[R_(∙)^(∙)_(mu nu)=],[del_(mu)Gamma_(nu∙)^(∙)-del_(nu)Gamma^(∙)_(mu∙)+[Gamma^(∙)_(mu∙),Gamma^(∙)_(nu∙)]","],[11.17)]:}\begin{aligned} & R_{\bullet}^{\bullet}{ }_{\mu \nu}= \\ & \partial_{\mu} \Gamma_{\nu \bullet}^{\bullet}-\partial_{\nu} \Gamma^{\bullet}{ }_{\mu \bullet}+\left[\Gamma^{\bullet}{ }_{\mu \bullet}, \Gamma^{\bullet}{ }_{\nu \bullet}\right], \\ & 11.17) \end{aligned}Rμν=μΓννΓμ+[Γμ,Γν],11.17)
where square brackets denote a commutator (i.e. [ A , B ] = A B B A [ A , B ] = A B B A [A,B]=AB-BA[A, B]=A B-B A[A,B]=ABBA ). If the connection coefficients are written out as matrices, eqn 11.19 has the additional advantage of simplifying the sums over indices. For example, in two dimensions we have
(11.18) Γ μ = ( Γ 1 μ 1 Γ 1 μ 2 Γ 2 μ 1 Γ 2 μ 2 ) (11.18) Γ μ = Γ 1 μ 1 Γ 1 μ 2 Γ 2 μ 1 Γ 2 μ 2 {:(11.18)Gamma_(mu∙)^(∙)=([Gamma^(1)_(mu1),Gamma^(1)_(mu2)],[Gamma^(2)_(mu1),Gamma^(2)_(mu2)]):}\Gamma_{\mu \bullet}^{\bullet}=\left(\begin{array}{cc} \Gamma^{1}{ }_{\mu 1} & \Gamma^{1}{ }_{\mu 2} \tag{11.18}\\ \Gamma^{2}{ }_{\mu 1} & \Gamma^{2}{ }_{\mu 2} \end{array}\right)(11.18)Γμ=(Γ1μ1Γ1μ2Γ2μ1Γ2μ2)
The matrix notation then allows the products of the connection coefficients to be computed using matrix algebra.
12 12 ^(12){ }^{12}12 Strictly we should write the final term on the right as 2 Φ x j x k g k i 2 Φ x j x k g k i (del^(2)Phi)/(delx^(j)delx^(k))g^(ki)\frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{k}} g^{k i}2Φxjxkgki if we were being fussy about tensor components.
13 13 ^(13){ }^{13}13 By 'symmetries' here we mean cases where components are, within a sign change, identical to other components. These mean we have to search for fewer distinct components of the tensor to characterize the curvature of spacetime. This is especially important for a ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor like R R R\boldsymbol{R}R, since is potentially has 4 4 = 256 4 4 = 256 4^(4)=2564^{4}=25644=256 different components in ( 3 + 1 ) ( 3 + 1 ) (3+1)(3+1)(3+1)-dimensional spacetime in the absence of symmetry.
14 14 ^(14){ }^{14}14 In a general coordinate frame, R R R\boldsymbol{R}R can be computed more directly as follows:
R α β μ ν = 1 2 ( g α ν , β μ g α μ , β ν + g β μ , α ν g β ν , α μ ) (11.22) + g σ ρ ( Γ σ β μ Γ ρ α ν Γ σ β ν Γ ρ α μ ) R α β μ ν = 1 2 g α ν , β μ g α μ , β ν + g β μ , α ν g β ν , α μ (11.22) + g σ ρ Γ σ β μ Γ ρ α ν Γ σ β ν Γ ρ α μ {:[R_(alpha beta mu nu)=],[(1)/(2)(g_(alpha nu,beta mu)-g_(alpha mu,beta nu):}],[{:+g_(beta mu,alpha nu)-g_(beta nu,alpha mu))],[(11.22)+g_(sigma rho)(Gamma^(sigma)_(beta mu)Gamma^(rho)_(alpha nu)-Gamma^(sigma)_(beta nu)Gamma^(rho)_(alpha mu))]:}\begin{align*} & R_{\alpha \beta \mu \nu}= \\ & \frac{1}{2}\left(g_{\alpha \nu, \beta \mu}-g_{\alpha \mu, \beta \nu}\right. \\ & \left.+g_{\beta \mu, \alpha \nu}-g_{\beta \nu, \alpha \mu}\right) \\ & +g_{\sigma \rho}\left(\Gamma^{\sigma}{ }_{\beta \mu} \Gamma^{\rho}{ }_{\alpha \nu}-\Gamma^{\sigma}{ }_{\beta \nu} \Gamma^{\rho}{ }_{\alpha \mu}\right) \tag{11.22} \end{align*}Rαβμν=12(gαν,βμgαμ,βν+gβμ,ανgβν,αμ)(11.22)+gσρ(ΓσβμΓρανΓσβνΓραμ)
We'll have little need to apply this expression in anger, but it's comforting to know that it exists. Note that the symmetries apply to both frames.
15 15 ^(15){ }^{15}15 The symmetries mean that R μ μ μ μ = R μ μ μ μ = R_(mu mu mu mu)=R_{\mu \mu \mu \mu}=Rμμμμ= R μ μ ν σ = R μ ν σ σ = 0 R μ μ ν σ = R μ ν σ σ = 0 R_(mu mu nu sigma)=R_(mu nu sigma sigma)=0R_{\mu \mu \nu \sigma}=R_{\mu \nu \sigma \sigma}=0Rμμνσ=Rμνσσ=0.
We stress finally that the curvature tensor is built from connection coefficients and the connection coefficients are build from the metric. The metric field is the source of curvature.

11.4 Symmetries of the Riemann tensor

The Riemann tensor is vitally important object in general relativity. The tensor has a number of important symmetries. 13 13 ^(13){ }^{13}13 Let's now find these. We'll consider the all-down index version of R R R\boldsymbol{R}R, achieved by saying R α β γ δ = g α μ R β γ δ μ R α β γ δ = g α μ R β γ δ μ R_(alpha beta gamma delta)=g_(alpha mu)R_(beta gamma delta)^(mu)R_{\alpha \beta \gamma \delta}=g_{\alpha \mu} R_{\beta \gamma \delta}^{\mu}Rαβγδ=gαμRβγδμ.

Example 11.5

The important local flatness theorem tells us it is always possible to find a local inertial frame (LIF) coordinate system such that the metric is identical to the Minkowski metric and the first derivative of the metric field vanishes at a point P P P\mathcal{P}P. We can therefore work at this point P P P\mathcal{P}P where, as a consequence, the connection vanishes such that Γ μ α β ( x = P ) = 0 Γ μ α β ( x = P ) = 0 Gamma^(mu)_(alpha beta)(x=P)=0\Gamma^{\mu}{ }_{\alpha \beta}(x=\mathcal{P})=0Γμαβ(x=P)=0. However, the derivatives of the connection do not vanish (since local flatness certainly does not imply the vanishing of the second derivatives of the metric field). Therefore, the Riemann tensor has components given by the derivatives metric field). Therefore, the Riemann tensor has components given by the derivatives
of the connections only. These can, in turn, be expressed as derivatives of the metric of the connections only. These can, in turn, be exp
components, with the result that, in the LIF, 14 14 ^(14){ }^{14}14
(11.23) R μ ^ α ^ α ^ β ^ = 1 2 ( 2 g α ^ ν ^ x β ^ x μ ^ 2 g α ^ μ ^ x β ^ x ν ^ + 2 g β ^ μ ^ x α ^ x ν ^ 2 g β ^ ν ^ x α ^ x μ ^ ) . (11.23) R μ ^ α ^ α ^ β ^ = 1 2 2 g α ^ ν ^ x β ^ x μ ^ 2 g α ^ μ ^ x β ^ x ν ^ + 2 g β ^ μ ^ x α ^ x ν ^ 2 g β ^ ν ^ x α ^ x μ ^ . {:(11.23)R_( hat(mu) hat(alpha) hat(alpha) hat(beta))=(1)/(2)((del^(2)g_( hat(alpha) hat(nu)))/(delx^( hat(beta))delx^( hat(mu)))-(del^(2)g_( hat(alpha) hat(mu)))/(delx^( hat(beta))delx^( hat(nu)))+(del^(2)g_( hat(beta) hat(mu)))/(delx^( hat(alpha))delx^( hat(nu)))-(del^(2)g_( hat(beta) hat(nu)))/(delx^( hat(alpha))delx^( hat(mu)))).:}\begin{equation*} R_{\hat{\mu} \hat{\alpha} \hat{\alpha} \hat{\beta}}=\frac{1}{2}\left(\frac{\partial^{2} g_{\hat{\alpha} \hat{\nu}}}{\partial x^{\hat{\beta}} \partial x^{\hat{\mu}}}-\frac{\partial^{2} g_{\hat{\alpha} \hat{\mu}}}{\partial x^{\hat{\beta}} \partial x^{\hat{\nu}}}+\frac{\partial^{2} g_{\hat{\beta} \hat{\mu}}}{\partial x^{\hat{\alpha}} \partial x^{\hat{\nu}}}-\frac{\partial^{2} g_{\hat{\beta} \hat{\nu}}}{\partial x^{\hat{\alpha}} \partial x^{\hat{\mu}}}\right) . \tag{11.23} \end{equation*}(11.23)Rμ^α^α^β^=12(2gα^ν^xβ^xμ^2gα^μ^xβ^xν^+2gβ^μ^xα^xν^2gβ^ν^xα^xμ^).
This equation, built from the symmetric metric tensor g g g\boldsymbol{g}g makes the symmetries of the curvature tensor R R R\boldsymbol{R}R manifest.
By considering eqn 11.23 we can identify the symmetries of R μ ν α β R μ ν α β R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}Rμναβ. Swapping indices within the first pair ( μ ν ) ( μ ν ) (mu nu)(\mu \nu)(μν) and second pair ( α β ) ( α β ) (alpha beta)(\alpha \beta)(αβ) results in picking up a minus sign, so we have 15 15 ^(15){ }^{15}15
R α β γ δ = R β α γ δ R α β γ δ = R α β δ γ (11.24) R α β γ δ = + R γ δ α β R α β γ δ = R β α γ δ R α β γ δ = R α β δ γ (11.24) R α β γ δ = + R γ δ α β {:[R_(alpha beta gamma delta)=-R_(beta alpha gamma delta)],[R_(alpha beta gamma delta)=-R_(alpha beta delta gamma)],[(11.24)R_(alpha beta gamma delta)=+R_(gamma delta alpha beta)]:}\begin{align*} R_{\alpha \beta \gamma \delta} & =-R_{\beta \alpha \gamma \delta} \\ R_{\alpha \beta \gamma \delta} & =-R_{\alpha \beta \delta \gamma} \\ R_{\alpha \beta \gamma \delta} & =+R_{\gamma \delta \alpha \beta} \tag{11.24} \end{align*}Rαβγδ=RβαγδRαβγδ=Rαβδγ(11.24)Rαβγδ=+Rγδαβ
In words, these say that swaps within the first or second pair of indices earn a minus sign; swapping the pairs together does not. We also can check that we have a cyclic identity
(11.25) R α β γ δ + R α δ β γ + R α γ δ β = 0 (11.25) R α β γ δ + R α δ β γ + R α γ δ β = 0 {:(11.25)R_(alpha beta gamma delta)+R_(alpha delta beta gamma)+R_(alpha gamma delta beta)=0:}\begin{equation*} R_{\alpha \beta \gamma \delta}+R_{\alpha \delta \beta \gamma}+R_{\alpha \gamma \delta \beta}=0 \tag{11.25} \end{equation*}(11.25)Rαβγδ+Rαδβγ+Rαγδβ=0
One consequence of these symmetries is on the makeup of the Riemann tensor. At first glance, it looks like the R R R\boldsymbol{R}R tensor in N N NNN dimensions has N 4 N 4 N^(4)N^{4}N4 components. However, these symmetries of the tensor strongly restrict the number of degrees of freedom, so that (as we will prove in Chapter 35) there are only N 2 ( N 2 1 ) / 12 N 2 N 2 1 / 12 N^(2)(N^(2)-1)//12N^{2}\left(N^{2}-1\right) / 12N2(N21)/12 independent components. We have for R μ ν α β R μ ν α β R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}Rμναβ that the number of independent components are as follows.
Dimensions 1 2 3 4
Independent components 0 1 6 20
Dimensions 1 2 3 4 Independent components 0 1 6 20| Dimensions | 1 | 2 | 3 | 4 | | :--- | :---: | :---: | :---: | :---: | | Independent components | 0 | 1 | 6 | 20 |
Some immediate consequences are considered in the following example.
Example 11.6
In one dimension, there is always a coordinate transformation that reduces the metric on a line to the Euclidean form. There is never any curvature. 16 16 ^(16){ }^{16}16
In two dimensions we have a single non-zero component of R R R\boldsymbol{R}R. According to the symmetries, this must be R 1212 R 1212 R_(1212)R_{1212}R1212

11.5 The Ricci tensor and Ricci scalar

Once we have the Riemann tensor R R R\boldsymbol{R}R and its symmetries then this also unlocks two simpler tensors that turn out to be essential to building the Einstein equation, linking the curvature and the matter fields of the Universe. It is perhaps to be expected that a ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor does not feature as raw material in a law of nature. Physics is full of ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) and ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) valent tensors, but very few ( 3 , 1 ) ( 3 , 1 ) (3,1)(3,1)(3,1) valent tensors. To build a ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor from a ( 3 , 1 ) ( 3 , 1 ) (3,1)(3,1)(3,1) tensor the simplest operation would be to contract the up index with a down index. 17 17 ^(17){ }^{17}17 The symmetries constrain the possible choices, with the result that we define the Ricci tensor 18 18 ^(18){ }^{18}18 as the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) valent tensor with components
R ν β = R μ ν μ β (11.26) = R 0 ν 0 β + R ν 1 β 1 + R ν 2 β 2 + R 3 ν 3 β . R ν β = R μ ν μ β (11.26) = R 0 ν 0 β + R ν 1 β 1 + R ν 2 β 2 + R 3 ν 3 β . {:[R_(nu beta)=R^(mu)_(nu mu beta)],[(11.26)=R^(0)_(nu0beta)+R_(nu1beta)^(1)+R_(nu2beta)^(2)+R^(3)_(nu3beta).]:}\begin{align*} R_{\nu \beta} & =R^{\mu}{ }_{\nu \mu \beta} \\ & =R^{0}{ }_{\nu 0 \beta}+R_{\nu 1 \beta}^{1}+R_{\nu 2 \beta}^{2}+R^{3}{ }_{\nu 3 \beta} . \tag{11.26} \end{align*}Rνβ=Rμνμβ(11.26)=R0ν0β+Rν1β1+Rν2β2+R3ν3β.
The symmetries of R R R\boldsymbol{R}R imply that the Ricci tensor is symmetric, such that R μ ν = R ν μ R μ ν = R ν μ R_(mu nu)=R_(nu mu)R_{\mu \nu}=R_{\nu \mu}Rμν=Rνμ. We can now make a further contraction, and turn the Ricci tensor into another tensor which is exceptionally simple: a ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) tensor, otherwise known of course as a scalar. We will discuss the resulting object in more detail in the following chapter, but let's define it now. It's called the Ricci scalar R R RRR and is obtained by tracing over the indices of the Ricci tensor. Hence, 19 19 ^(19){ }^{19}19 the Ricci scalar R R RRR is given by
(11.27) R = g μ ν R μ ν (11.27) R = g μ ν R μ ν {:(11.27)R=g^(mu nu)R_(mu nu):}\begin{equation*} R=g^{\mu \nu} R_{\mu \nu} \tag{11.27} \end{equation*}(11.27)R=gμνRμν
The significance of the Ricci tensor will become clear as we try to relate this measure of curvature to the mass-energy of the fields filling the Universe. For now, you can think of it as a kind of average over components of the Riemann tensor. 20 20 ^(20){ }^{20}20
16 16 ^(16){ }^{16}16 The one-dimensional example is an interesting, but trivial, case. It has no intrinsic curvature. You can of course embed a one-dimensional space in a higher dimension, and make it appear curved by tying it into a bow or winding it around a cylinder, but that curvature is always extrinsic and never intrinsic. The one-dimensional space is always isometric to a perfectly straight line. Things are more complicated in line. Things are more complicated in
two dimensions; an ant confined to the two dimensions; an ant confined to the
surface of a large ball only knows it is living in a curved space if it goes for a walk around the surface and measures how a vector is parallel transported around its walk and compares the result on returning to its initial position. Such closed-loop walks are not possible in one dimension.
17 A 17 A ^(17)A{ }^{17} \mathrm{~A}17 A contraction is an operation on a tensor that involves setting an upstairs index equal to a downstairs one and summing. The result is to reduce the number of indices by two.
18 18 ^(18){ }^{18}18 Gregorio Ricci-Curbastro (18531925). The Ricci tensor is one of the few important tensors to which we do not assign a bold symbol. The reason is that it does not appear in the field theory of gravity alone, but in a so-called trace-reversed form R μ ν 1 2 g μ ν R R μ ν 1 2 g μ ν R R_(mu nu)-(1)/(2)g_(mu nu)RR_{\mu \nu}-\frac{1}{2} g_{\mu \nu} RRμν12gμνR, where R R RRR is the Ricci scalar (the trace over R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν, see below). These are the components of the Einstein tensor G G G\boldsymbol{G}G, which is the quantity given a bold symbol. Some texts do not follow this convention and assign the Ricci tensor a bold symbol such as Ri.
19 A 19 A ^(19)A{ }^{19} \mathrm{~A}19 A trace is often written as A μ μ = A μ μ = A^(mu)_(mu)=A^{\mu}{ }_{\mu}=Aμμ= A 0 0 + A 1 1 + A 2 2 + A 3 3 A 0 0 + A 1 1 + A 2 2 + A 3 3 A^(0)_(0)+A^(1)_(1)+A^(2)_(2)+A^(3)_(3)A^{0}{ }_{0}+A^{1}{ }_{1}+A^{2}{ }_{2}+A^{3}{ }_{3}A00+A11+A22+A33. In general relativity, it is best written g μ ν A μ ν g μ ν A μ ν g_(mu nu)A^(mu nu)g_{\mu \nu} A^{\mu \nu}gμνAμν as an aide-memoir that the metric must be used to construct the trace in curved spacetime.
20 20 ^(20){ }^{20}20 It will turn out that it is this average that captures the ability of gravitational curvature to cause volumes to shrink.

11.6 Example computations

We shall now carry out two brute force (and ignorance) calculations of the components of the Riemann tensor, using the equation
(11.28) R ν α β μ = Γ β ν μ x α Γ α ν μ x β + Γ α ρ μ Γ β ν ρ Γ β ρ μ β Γ α ν ρ . (11.28) R ν α β μ = Γ β ν μ x α Γ α ν μ x β + Γ α ρ μ Γ β ν ρ Γ β ρ μ β Γ α ν ρ . {:(11.28)R_(nu alpha beta)^(mu)=(delGamma_(beta nu)^(mu))/(delx^(alpha))-(delGamma_(alpha nu)^(mu))/(delx^(beta))+Gamma_(alpha rho)^(mu)Gamma_(beta nu)^(rho)-Gamma_(beta rho)^(mu)_(beta)Gamma_(alpha nu)^(rho).:}\begin{equation*} R_{\nu \alpha \beta}^{\mu}=\frac{\partial \Gamma_{\beta \nu}^{\mu}}{\partial x^{\alpha}}-\frac{\partial \Gamma_{\alpha \nu}^{\mu}}{\partial x^{\beta}}+\Gamma_{\alpha \rho}^{\mu} \Gamma_{\beta \nu}^{\rho}-\Gamma_{\beta \rho}^{\mu}{ }_{\beta} \Gamma_{\alpha \nu}^{\rho} . \tag{11.28} \end{equation*}(11.28)Rναβμ=ΓβνμxαΓανμxβ+ΓαρμΓβνρΓβρμβΓανρ.
Example 11.7
First consider flat space with line element d s 2 = d r 2 + r 2 d θ 2 d s 2 = d r 2 + r 2 d θ 2 ds^(2)=dr^(2)+r^(2)dtheta^(2)\mathrm{d} s^{2}=\mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2}ds2=dr2+r2 dθ2. We expect that all of the components of the Riemann tensor should vanish. We saw that this metric has connections Γ r θ θ = r Γ r θ θ = r Gamma^(r)_(theta theta)=-r\Gamma^{r}{ }_{\theta \theta}=-rΓrθθ=r and Γ θ r θ = 1 r Γ θ r θ = 1 r Gamma^(theta)_(r theta)=(1)/(r)\Gamma^{\theta}{ }_{r \theta}=\frac{1}{r}Γθrθ=1r. We know from the symmetries that R r r r r = R θ θ θ θ = 0 R r r r r = R θ θ θ θ = 0 R_(rrr)^(r)=R_(theta theta theta)^(theta)=0R_{r r r}^{r}=R_{\theta \theta \theta}^{\theta}=0Rrrrr=Rθθθθ=0. Let's try some of the mixed-index components
R r θ r θ = Γ θ r r x θ Γ θ θ r x r + Γ θ θ ρ Γ ρ r r Γ r ρ θ Γ θ r ρ (11.29) = 0 + 1 r 2 + 0 1 r 1 r = 0 . R r θ r θ = Γ θ r r x θ Γ θ θ r x r + Γ θ θ ρ Γ ρ r r Γ r ρ θ Γ θ r ρ (11.29) = 0 + 1 r 2 + 0 1 r 1 r = 0 . {:[R_(r theta r)^(theta)=(delGamma^(theta)_(rr))/(delx^(theta))-(delGamma^(theta)_(theta r))/(delx^(r))+Gamma^(theta)_(theta rho)Gamma^(rho)_(rr)-Gamma_(r rho)^(theta)Gamma_(theta r)^(rho)],[(11.29)=0+(1)/(r^(2))+0-(1)/(r)(1)/(r)=0.]:}\begin{align*} R_{r \theta r}^{\theta} & =\frac{\partial \Gamma^{\theta}{ }_{r r}}{\partial x^{\theta}}-\frac{\partial \Gamma^{\theta}{ }_{\theta r}}{\partial x^{r}}+\Gamma^{\theta}{ }_{\theta \rho} \Gamma^{\rho}{ }_{r r}-\Gamma_{r \rho}^{\theta} \Gamma_{\theta r}^{\rho} \\ & =0+\frac{1}{r^{2}}+0-\frac{1}{r} \frac{1}{r}=0 . \tag{11.29} \end{align*}Rrθrθ=ΓθrrxθΓθθrxr+ΓθθρΓρrrΓrρθΓθrρ(11.29)=0+1r2+01r1r=0.
Another one is
R θ r θ r = Γ θ θ r x r Γ r θ r x θ + Γ r ρ r Γ θ θ ρ Γ θ ρ r Γ r θ ρ R θ r θ r = Γ θ θ r x r Γ r θ r x θ + Γ r ρ r Γ θ θ ρ Γ θ ρ r Γ r θ ρ R_(theta r theta)^(r)=(delGamma_(theta theta)^(r))/(delx^(r))-(delGamma_(r theta)^(r))/(delx^(theta))+Gamma_(r rho)^(r)Gamma_(theta theta)^(rho)-Gamma_(theta rho)^(r)Gamma_(r theta)^(rho)R_{\theta r \theta}^{r}=\frac{\partial \Gamma_{\theta \theta}^{r}}{\partial x^{r}}-\frac{\partial \Gamma_{r \theta}^{r}}{\partial x^{\theta}}+\Gamma_{r \rho}^{r} \Gamma_{\theta \theta}^{\rho}-\Gamma_{\theta \rho}^{r} \Gamma_{r \theta}^{\rho}Rθrθr=ΓθθrxrΓrθrxθ+ΓrρrΓθθρΓθρrΓrθρ
(11.30) = 1 + 0 + 0 + 1 = 0 . (11.30) = 1 + 0 + 0 + 1 = 0 . {:(11.30)=-1+0+0+1=0.:}\begin{equation*} =-1+0+0+1=0 . \tag{11.30} \end{equation*}(11.30)=1+0+0+1=0.
21 21 ^(21){ }^{21}21 Consequently, the Ricci tensor and Ricci scalar both vanish. This example may not have seemed that exciting, but it's good to check that our formalism works. Flat space is indeed not curved.
As expected, the components all vanish in flat space. 21 21 ^(21){ }^{21}21
Now the (considerably less dull) example of a two-dimensional spherical space on the surface of a sphere of radius a a aaa.

Example 11.8

The spherical space has a line element
(11.31) d s 2 = a 2 d θ 2 + a 2 sin 2 θ d ϕ 2 . (11.31) d s 2 = a 2 d θ 2 + a 2 sin 2 θ d ϕ 2 . {:(11.31)ds^(2)=a^(2)dtheta^(2)+a^(2)sin^(2)thetadphi^(2).:}\begin{equation*} \mathrm{d} s^{2}=a^{2} \mathrm{~d} \theta^{2}+a^{2} \sin ^{2} \theta \mathrm{~d} \phi^{2} . \tag{11.31} \end{equation*}(11.31)ds2=a2 dθ2+a2sin2θ dϕ2.
This gives connection coefficients
(11.32) Γ ϕ ϕ θ = sin θ cos θ , Γ θ ϕ ϕ = cot θ . (11.32) Γ ϕ ϕ θ = sin θ cos θ , Γ θ ϕ ϕ = cot θ . {:(11.32)Gamma_(phi phi)^(theta)=-sin theta cos theta","quadGamma_(theta phi)^(phi)=cot theta.:}\begin{equation*} \Gamma_{\phi \phi}^{\theta}=-\sin \theta \cos \theta, \quad \Gamma_{\theta \phi}^{\phi}=\cot \theta . \tag{11.32} \end{equation*}(11.32)Γϕϕθ=sinθcosθ,Γθϕϕ=cotθ.
The components may be calculated. Here's an example
(11.33) R ϕ θ ϕ θ = Γ ϕ ϕ θ x θ . Γ θ θ ϕ x ϕ + Γ θ θ ρ Γ ρ ϕ ϕ Γ θ ϕ ρ Γ ρ θ ϕ = sin 2 θ cos 2 θ 0 + 0 + cos 2 θ = sin 2 θ . (11.33) R ϕ θ ϕ θ = Γ ϕ ϕ θ x θ . Γ θ θ ϕ x ϕ + Γ θ θ ρ Γ ρ ϕ ϕ Γ θ ϕ ρ Γ ρ θ ϕ = sin 2 θ cos 2 θ 0 + 0 + cos 2 θ = sin 2 θ . {:[(11.33)R_(phi theta phi)^(theta)=(delGamma_(phi phi)^(theta))/(delx^(theta)).-(delGamma^(theta)_(theta phi))/(delx^(phi))+Gamma^(theta)_(theta rho)Gamma^(rho)_(phi phi)-Gamma^(theta)_(phi rho)Gamma^(rho)_(theta phi)],[=sin^(2)theta-cos^(2)theta-0+0+cos^(2)theta=sin^(2)theta.]:}\begin{align*} R_{\phi \theta \phi}^{\theta} & =\frac{\partial \Gamma_{\phi \phi}^{\theta}}{\partial x^{\theta}} .-\frac{\partial \Gamma^{\theta}{ }_{\theta \phi}}{\partial x^{\phi}}+\Gamma^{\theta}{ }_{\theta \rho} \Gamma^{\rho}{ }_{\phi \phi}-\Gamma^{\theta}{ }_{\phi \rho} \Gamma^{\rho}{ }_{\theta \phi} \tag{11.33}\\ & =\sin ^{2} \theta-\cos ^{2} \theta-0+0+\cos ^{2} \theta=\sin ^{2} \theta . \end{align*}(11.33)Rϕθϕθ=Γϕϕθxθ.Γθθϕxϕ+ΓθθρΓρϕϕΓθϕρΓρθϕ=sin2θcos2θ0+0+cos2θ=sin2θ.
Another one is 22 22 ^(22){ }^{22}22
R θ ϕ θ ϕ = Γ θ θ ϕ x ϕ Γ ϕ θ ϕ x θ + Γ ϕ ρ ϕ Γ θ θ ρ Γ θ ρ ϕ Γ ϕ θ ρ R θ ϕ θ ϕ = Γ θ θ ϕ x ϕ Γ ϕ θ ϕ x θ + Γ ϕ ρ ϕ Γ θ θ ρ Γ θ ρ ϕ Γ ϕ θ ρ R_(theta phi theta)^(phi)=(delGamma_(theta theta)^(phi))/(delx^(phi))-(delGamma_(phi theta)^(phi))/(delx^(theta))+Gamma_(phi rho)^(phi)Gamma_(theta theta)^(rho)-Gamma_(theta rho)^(phi)Gamma_(phi theta)^(rho)R_{\theta \phi \theta}^{\phi}=\frac{\partial \Gamma_{\theta \theta}^{\phi}}{\partial x^{\phi}}-\frac{\partial \Gamma_{\phi \theta}^{\phi}}{\partial x^{\theta}}+\Gamma_{\phi \rho}^{\phi} \Gamma_{\theta \theta}^{\rho}-\Gamma_{\theta \rho}^{\phi} \Gamma_{\phi \theta}^{\rho}Rθϕθϕ=ΓθθϕxϕΓϕθϕxθ+ΓϕρϕΓθθρΓθρϕΓϕθρ
(11.34) = 0 + 1 sin 2 θ + 0 cos 2 θ sin 2 θ = 1 (11.34) = 0 + 1 sin 2 θ + 0 cos 2 θ sin 2 θ = 1 {:(11.34)=0+(1)/(sin^(2)theta)+0-(cos^(2)theta)/(sin^(2)theta)=1:}\begin{equation*} =0+\frac{1}{\sin ^{2} \theta}+0-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}=1 \tag{11.34} \end{equation*}(11.34)=0+1sin2θ+0cos2θsin2θ=1
We have for the components of the Ricci tensor that
(11.35) R ϕ ϕ = sin 2 θ , R θ θ = 1 (11.35) R ϕ ϕ = sin 2 θ , R θ θ = 1 {:(11.35)R_(phi phi)=sin^(2)theta","quadR_(theta theta)=1:}\begin{equation*} R_{\phi \phi}=\sin ^{2} \theta, \quad R_{\theta \theta}=1 \tag{11.35} \end{equation*}(11.35)Rϕϕ=sin2θ,Rθθ=1
and the Ricci scalar is, therefore,
R = g θ θ R θ θ + g ϕ ϕ R ϕ ϕ (11.36) = 1 a 2 + sin 2 θ a 2 sin 2 θ = 2 a 2 . R = g θ θ R θ θ + g ϕ ϕ R ϕ ϕ (11.36) = 1 a 2 + sin 2 θ a 2 sin 2 θ = 2 a 2 . {:[R=g^(theta theta)R_(theta theta)+g^(phi phi)R_(phi phi)],[(11.36)=(1)/(a^(2))+(sin^(2)theta)/(a^(2)sin^(2)theta)=(2)/(a^(2)).]:}\begin{align*} R & =g^{\theta \theta} R_{\theta \theta}+g^{\phi \phi} R_{\phi \phi} \\ & =\frac{1}{a^{2}}+\frac{\sin ^{2} \theta}{a^{2} \sin ^{2} \theta}=\frac{2}{a^{2}} . \tag{11.36} \end{align*}R=gθθRθθ+gϕϕRϕϕ(11.36)=1a2+sin2θa2sin2θ=2a2.
These two examples show how the components of Riemann tensor R R R\boldsymbol{R}R can be evaluated via direct (admittedly, rather tedious) computation. Fortunately, there is a more efficient method to achieve this, although it does rely on a little more geometrical technology, as we'll see in Part V.

11.7 Geodesic deviation revisited

We now have access to the tensor R R R\boldsymbol{R}R that allows us to evaluate the curvature of spacetime. However, as we saw in Chapter 10, our measurements are made in orthonormal frames. By employing a specially-chosen orthonormal frame we can link the components of the Riemann tensor more directly to the notion of geodesic deviation we discussed earlier. The local frame we will choose is the freely falling local inertial frame.
We imagine setting a swarm of particles free from rest and allowing them to follow their geodesics. We assign one the status of being the fiducial geodesic and note its tangent u u u\boldsymbol{u}u. If the particles accelerate relative to each other then we can compute R R R\boldsymbol{R}R. We therefore need to work out the acceleration of a vector ξ ξ xi\boldsymbol{\xi}ξ linking a neighbouring geodesic to the fiducial one, which is related to the Riemann tensor via 23 23 ^(23){ }^{23}23
(11.38) D 2 ξ d τ 2 = R ( , u , ξ , u ) (11.38) D 2 ξ d τ 2 = R ( , u , ξ , u ) {:(11.38)(D^(2)xi)/((d)tau^(2))=R(","u","xi","u):}\begin{equation*} \frac{D^{2} \boldsymbol{\xi}}{\mathrm{~d} \tau^{2}}=\boldsymbol{R}(, \boldsymbol{u}, \boldsymbol{\xi}, \boldsymbol{u}) \tag{11.38} \end{equation*}(11.38)D2ξ dτ2=R(,u,ξ,u)
or, in components
(11.39) ( D 2 ξ d τ 2 ) α = R β γ δ α u β ξ γ u δ . (11.39) D 2 ξ d τ 2 α = R β γ δ α u β ξ γ u δ . {:(11.39)((D^(2)xi)/((d)tau^(2)))^(alpha)=-R_(beta gamma delta)^(alpha)u^(beta)xi^(gamma)u^(delta).:}\begin{equation*} \left(\frac{D^{2} \xi}{\mathrm{~d} \tau^{2}}\right)^{\alpha}=-R_{\beta \gamma \delta}^{\alpha} u^{\beta} \xi^{\gamma} u^{\delta} . \tag{11.39} \end{equation*}(11.39)(D2ξ dτ2)α=Rβγδαuβξγuδ.
In the coordinates of a freely falling frame we have 24 u β ^ = ( 1 , 0 , 0 , 0 ) 24 u β ^ = ( 1 , 0 , 0 , 0 ) ^(24)u^( hat(beta))=(1,0,0,0){ }^{24} u^{\hat{\beta}}=(1,0,0,0)24uβ^=(1,0,0,0), the geodesic deviation equation simplifies to 25 25 ^(25){ }^{25}25
(11.40) d 2 ξ α ^ d τ 2 = R α ^ t ^ β ^ t ^ ξ β ^ . (11.40) d 2 ξ α ^ d τ 2 = R α ^ t ^ β ^ t ^ ξ β ^ . {:(11.40)(d^(2)xi^( hat(alpha)))/(dtau^(2))=-R^( hat(alpha))_( hat(t) hat(beta) hat(t))xi^( hat(beta)).:}\begin{equation*} \frac{\mathrm{d}^{2} \xi^{\hat{\alpha}}}{\mathrm{d} \tau^{2}}=-R^{\hat{\alpha}}{ }_{\hat{t} \hat{\beta} \hat{t}} \xi^{\hat{\beta}} . \tag{11.40} \end{equation*}(11.40)d2ξα^dτ2=Rα^t^β^t^ξβ^.
Using the vielbein, we relate the coordinates of R R R\boldsymbol{R}R in the local freely falling frame to a coordinate frame via R α ^ β ^ γ ^ δ ^ = R α ^ β ^ γ ^ δ ^ = R^( hat(alpha))_( hat(beta) hat(gamma) hat(delta))=R^{\hat{\alpha}}{ }_{\hat{\beta} \hat{\gamma} \hat{\delta}}=Rα^β^γ^δ^= R μ ν λ σ ( e μ ) α ^ ( e β ^ ) ν ( e γ ^ ) λ ( e δ ^ ) σ R μ ν λ σ e μ α ^ e β ^ ν e γ ^ λ e δ ^ σ R^(mu)_(nu lambda sigma)(e_(mu))^( hat(alpha))(e_( hat(beta)))^(nu)(e_( hat(gamma)))^(lambda)(e_( hat(delta)))^(sigma)R^{\mu}{ }_{\nu \lambda \sigma}\left(\boldsymbol{e}_{\mu}\right)^{\hat{\alpha}}\left(\boldsymbol{e}_{\hat{\beta}}\right)^{\nu}\left(\boldsymbol{e}_{\hat{\gamma}}\right)^{\lambda}\left(\boldsymbol{e}_{\hat{\delta}}\right)^{\sigma}Rμνλσ(eμ)α^(eβ^)ν(eγ^)λ(eδ^)σ. We can use the simplified form of the deviation equation in eqn 11.40 to investigate a simple example.

Example 11.9

An important example is the Schwarzschild geometry of a spherically symmetric gravitating object. 26 26 ^(26){ }^{26}26 As we shall see, the metric for this spacetime gives rise to a Riemann curvature tensor R R R\boldsymbol{R}R with components in the freely falling frame of
(11.41) R t ^ t ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R t ^ θ ^ t ^ θ ^ = R t ^ ϕ ^ t ^ ϕ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ r ^ ϕ ^ = M r 3 (11.41) R t ^ t ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R t ^ θ ^ t ^ θ ^ = R t ^ ϕ ^ t ^ ϕ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ r ^ ϕ ^ = M r 3 {:(11.41){:[R_( hat(t) hat(t) hat(r))=-(2M)/(r^(3))",",R_( hat(theta) hat(phi) hat(theta) hat(phi))=+(2M)/(r^(3))","],[R_( hat(t) hat(theta) hat(t) hat(theta))=R_( hat(t) hat(phi) hat(t) hat(phi))=+(M)/(r^(3))",",R_( hat(r) hat(theta) hat(r) hat(theta))=R_( hat(r) hat(phi) hat(r) hat(phi))=-(M)/(r^(3))]:}:}\begin{array}{cc} R_{\hat{t} \hat{t} \hat{r}}=-\frac{2 M}{r^{3}}, & R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=+\frac{2 M}{r^{3}}, \\ R_{\hat{t} \hat{\theta} \hat{t} \hat{\theta}}=R_{\hat{t} \hat{\phi} \hat{t} \hat{\phi}}=+\frac{M}{r^{3}}, & R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=R_{\hat{r} \hat{\phi} \hat{r} \hat{\phi}}=-\frac{M}{r^{3}} \tag{11.41} \end{array}(11.41)Rt^t^r^=2Mr3,Rθ^ϕ^θ^ϕ^=+2Mr3,Rt^θ^t^θ^=Rt^ϕ^t^ϕ^=+Mr3,Rr^θ^r^θ^=Rr^ϕ^r^ϕ^=Mr3
Since we're in an orthonormal frame, we use the Minkowski tensor to raise indices and the geodesic deviation equations become
(11.42) d 2 ξ ^ d τ 2 = 2 M r 3 ξ r ^ , d 2 ξ θ ^ d τ 2 = M r 3 ξ θ ^ , d 2 ξ β ^ d τ 2 = M r 3 ξ ϕ ^ (11.42) d 2 ξ ^ d τ 2 = 2 M r 3 ξ r ^ , d 2 ξ θ ^ d τ 2 = M r 3 ξ θ ^ , d 2 ξ β ^ d τ 2 = M r 3 ξ ϕ ^ {:(11.42)(d^(2)xi^( hat()))/(dtau^(2))=(2M)/(r^(3))xi^( hat(r))","quad(d^(2)xi^( hat(theta)))/(dtau^(2))=-(M)/(r^(3))xi^( hat(theta))","quad(d^(2)xi^( hat(beta)))/(dtau^(2))=-(M)/(r^(3))xi^( hat(phi)):}\begin{equation*} \frac{\mathrm{d}^{2} \xi^{\hat{}}}{\mathrm{d} \tau^{2}}=\frac{2 M}{r^{3}} \xi^{\hat{r}}, \quad \frac{\mathrm{~d}^{2} \xi^{\hat{\theta}}}{\mathrm{d} \tau^{2}}=-\frac{M}{r^{3}} \xi^{\hat{\theta}}, \quad \frac{\mathrm{d}^{2} \xi^{\hat{\beta}}}{\mathrm{d} \tau^{2}}=-\frac{M}{r^{3}} \xi^{\hat{\phi}} \tag{11.42} \end{equation*}(11.42)d2ξ^dτ2=2Mr3ξr^, d2ξθ^dτ2=Mr3ξθ^,d2ξβ^dτ2=Mr3ξϕ^
23 23 ^(23){ }^{23}23 This expression is sometimes called the Jacobi equation. The double derivative can be written as
(11.37) D 2 ξ d τ 2 = u u ξ (11.37) D 2 ξ d τ 2 = u u ξ {:(11.37)(D^(2)xi)/((d)tau^(2))=grad_(u)grad_(u)xi:}\begin{equation*} \frac{D^{2} \boldsymbol{\xi}}{\mathrm{~d} \tau^{2}}=\nabla_{u} \nabla_{u} \boldsymbol{\xi} \tag{11.37} \end{equation*}(11.37)D2ξ dτ2=uuξ
It is also useful to recall that in an orthonormal frame we take u = e i ^ u = e i ^ u=e_( hat(i))\boldsymbol{u}=\boldsymbol{e}_{\hat{i}}u=ei^.
24 24 ^(24){ }^{24}24 That is, the observer is at rest and so the only non-zero component of u u u\boldsymbol{u}u is u t ^ u t ^ u^( hat(t))u^{\hat{t}}ut^. Since in the local coordinate system the observer uses the Minkowski tensor η η eta\boldsymbol{\eta}η to form products, the normalization u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1 becomes η t ^ t ^ ( u t ^ ) 2 = 1 η t ^ t ^ u t ^ 2 = 1 eta_( hat(t) hat(t))(u^( hat(t)))^(2)=-1\eta_{\hat{t} \hat{t}}\left(u^{\hat{t}}\right)^{2}=-1ηt^t^(ut^)2=1 and so, since η t ^ t ^ = 1 η t ^ t ^ = 1 eta_( hat(t) hat(t))=-1\eta_{\hat{t} \hat{t}}=-1ηt^t^=1, we have u t ^ = 1 u t ^ = 1 u^( hat(t))=1u^{\hat{t}}=1ut^=1.
25 25 ^(25){ }^{25}25 The use of the freely falling frame allows the swap ( D 2 ξ / d τ 2 ) α D 2 ξ / d τ 2 α (D^(2)xi//dtau^(2))^(alpha)\left(\mathrm{D}^{2} \boldsymbol{\xi} / \mathrm{d} \tau^{2}\right)^{\alpha}(D2ξ/dτ2)α to d 2 ξ α / d τ 2 d 2 ξ α / d τ 2 d^(2)xi^(alpha)//dtau^(2)\mathrm{d}^{2} \xi^{\alpha} / \mathrm{d} \tau^{2}d2ξα/dτ2. Notice how the resulting expression is analogous to a simple harmonic oscillator: x ¨ = ω 0 2 x x ¨ = ω 0 2 x x^(¨)=-omega_(0)^(2)x\ddot{x}=-\omega_{0}^{2} xx¨=ω02x (and it is identical to this equation of motion for α ^ = β ^ α ^ = β ^ hat(alpha)= hat(beta)\hat{\alpha}=\hat{\beta}α^=β^ ). The analogue of the natural frequency of the oscillator is provided by the components of the Riemann tensor. In this way, the curvature of spacetime can be thought of as analogous to a spring constant. 26 26 ^(26){ }^{26}26 Discussed in Part IV.
Notice the resemblance of these equations of motion to those of the Newtonian problem discussed in Example 11.1.

Chapter summary

  • Geodesic deviation allows us to identify tidal forces that result from the curvature of spacetime due to gravity.
  • The Riemann tensor R R R\boldsymbol{R}R is non-zero if there is intrinsic curvature in spacetime.
  • The tensor R R R\boldsymbol{R}R can be derived by parallel transporting a vector around a loop. The components of R R R\boldsymbol{R}R are given by
(11.43) R α β μ ν = [ Γ α ν β x μ Γ α μ β x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ ] . (11.43) R α β μ ν = Γ α ν β x μ Γ α μ β x ν + Γ μ σ α Γ ν β σ Γ ν σ α Γ μ β σ . {:(11.43)R^(alpha)_(beta mu nu)=[(delGamma^(alpha)_(nu beta))/(delx^(mu))-(delGamma^(alpha)_(mu beta))/(delx^(nu))+Gamma_(mu sigma)^(alpha)Gamma_(nu beta)^(sigma)-Gamma_(nu sigma)^(alpha)Gamma_(mu beta)^(sigma)].:}\begin{equation*} R^{\alpha}{ }_{\beta \mu \nu}=\left[\frac{\partial \Gamma^{\alpha}{ }_{\nu \beta}}{\partial x^{\mu}}-\frac{\partial \Gamma^{\alpha}{ }_{\mu \beta}}{\partial x^{\nu}}+\Gamma_{\mu \sigma}^{\alpha} \Gamma_{\nu \beta}^{\sigma}-\Gamma_{\nu \sigma}^{\alpha} \Gamma_{\mu \beta}^{\sigma}\right] . \tag{11.43} \end{equation*}(11.43)Rαβμν=[ΓανβxμΓαμβxν+ΓμσαΓνβσΓνσαΓμβσ].

Exercises

(11.1) Using the connection coefficients computed for the Schwarzschild geometry in Exercise 9.7, show the following:
R r t r t = Φ + Φ Λ ( Φ ) 2 , R θ t θ t = r e 2 Λ Φ R ϕ t ϕ t = r e 2 Λ sin 2 θ Φ , R θ r θ r = r e 2 Λ Λ R ϕ r ϕ r = r e 2 Λ Λ sin 2 θ , (11.44) R ϕ θ ϕ θ = ( 1 e 2 Λ ) sin 2 θ . R r t r t = Φ + Φ Λ Φ 2 , R θ t θ t = r e 2 Λ Φ R ϕ t ϕ t = r e 2 Λ sin 2 θ Φ , R θ r θ r = r e 2 Λ Λ R ϕ r ϕ r = r e 2 Λ Λ sin 2 θ , (11.44) R ϕ θ ϕ θ = 1 e 2 Λ sin 2 θ . {:[R_(rtr)^(t)=-Phi^('')+Phi^(')Lambda^(')-(Phi^('))^(2)","],[R_(theta t theta)^(t)=-re^(-2Lambda)Phi^(')],[R_(phi t phi)^(t)=-re^(-2Lambda)sin^(2)thetaPhi^(')","],[R_(theta r theta)^(r)=re^(-2Lambda)Lambda^(')],[R_(phi r phi)^(r)=re^(-2Lambda)Lambda^(')sin^(2)theta","],[(11.44)R_(phi theta phi)^(theta)=(1-e^(-2Lambda))sin^(2)theta.]:}\begin{align*} R_{r t r}^{t} & =-\Phi^{\prime \prime}+\Phi^{\prime} \Lambda^{\prime}-\left(\Phi^{\prime}\right)^{2}, \\ R_{\theta t \theta}^{t} & =-r \mathrm{e}^{-2 \Lambda} \Phi^{\prime} \\ R_{\phi t \phi}^{t} & =-r \mathrm{e}^{-2 \Lambda} \sin ^{2} \theta \Phi^{\prime}, \\ R_{\theta r \theta}^{r} & =r \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \\ R_{\phi r \phi}^{r} & =r \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \sin ^{2} \theta, \\ R_{\phi \theta \phi}^{\theta} & =\left(1-\mathrm{e}^{-2 \Lambda}\right) \sin ^{2} \theta . \tag{11.44} \end{align*}Rrtrt=Φ+ΦΛ(Φ)2,Rθtθt=re2ΛΦRϕtϕt=re2Λsin2θΦ,Rθrθr=re2ΛΛRϕrϕr=re2ΛΛsin2θ,(11.44)Rϕθϕθ=(1e2Λ)sin2θ.
(11.2) We shall see in Chapter 21 that the Schwarzschild metric line element outside a star of mass M M MMM can be written as
d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)],[+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{aligned} \mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \end{aligned}ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
In terms of these variables, the components of R R R\boldsymbol{R}R from the last problem can be written as
R r t r t = 2 M r 2 ( 2 M r ) , R θ t θ t = M r R ϕ t ϕ t = M sin 2 θ r R θ r θ r = M r R ϕ r ϕ r = M sin 2 θ r (11.46) R ϕ θ ϕ θ = 2 M sin 2 θ r R r t r t = 2 M r 2 ( 2 M r ) , R θ t θ t = M r R ϕ t ϕ t = M sin 2 θ r R θ r θ r = M r R ϕ r ϕ r = M sin 2 θ r (11.46) R ϕ θ ϕ θ = 2 M sin 2 θ r {:[R_(rtr)^(t)=-(2M)/(r^(2)(2M-r))","],[R_(theta t theta)^(t)=-(M)/(r)],[R_(phi t phi)^(t)=-(Msin^(2)theta)/(r)],[R_(theta r theta)^(r)=-(M)/(r)],[R_(phi r phi)^(r)=-(Msin^(2)theta)/(r)],[(11.46)R_(phi theta phi)^(theta)=(2Msin^(2)theta)/(r)]:}\begin{align*} R_{r t r}^{t} & =-\frac{2 M}{r^{2}(2 M-r)}, \\ R_{\theta t \theta}^{t} & =-\frac{M}{r} \\ R_{\phi t \phi}^{t} & =-\frac{M \sin ^{2} \theta}{r} \\ R_{\theta r \theta}^{r} & =-\frac{M}{r} \\ R_{\phi r \phi}^{r} & =-\frac{M \sin ^{2} \theta}{r} \\ R_{\phi \theta \phi}^{\theta} & =\frac{2 M \sin ^{2} \theta}{r} \tag{11.46} \end{align*}Rrtrt=2Mr2(2Mr),Rθtθt=MrRϕtϕt=Msin2θrRθrθr=MrRϕrϕr=Msin2θr(11.46)Rϕθϕθ=2Msin2θr
(a) Write expressions for (i) R r r t t R r r t t R_(rrt)^(t)R_{r r t}^{t}Rrrtt and (ii) R t r t r R t r t r R_(trt)^(r)R_{t r t}^{r}Rtrtr.
(b) Compute the components of R R R\boldsymbol{R}R in the orthonormal frame.
(11.3) Using the components in Exercise 11.2, show that the components of the Ricci tensor vanish for the Schwarzschild metric outside a star. Why must this be the case?
(11.4) Demonstrate the spacetime described by the rotating frame metric in Exercise 3.5 is flat.

The energy-momentum tensor

The curvature that we discussed in the last chapter provides the lefthand, geometrical side of Einstein's field equation. The right-hand, physical side is supplied by the mass-energy content of the matter fields that fill time and space. 1 1 ^(1){ }^{1}1 The mass-energy content of spacetime is expressed using the energy-momentum tensor field T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x). We met an example of the energy-momentum tensor briefly in Part I of the book. In this chapter, we meet the tensor again in a little more detail, along with its all-important constraint T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. These tools will enable us, in the next chapter, to put everything together and write down the Einstein field equation for general relativity.

12.1 Another look at the energy-momentum tensor

The energy-momentum tensor field T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x) gives us access to a tensor T ( T ( T(\boldsymbol{T}(T(, at each point in spacetime that can be used to evaluate the energy and momentum content of matter fields at that point. If in doubt, you can think of T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x) as describing the current of momentum for a distribution of mass-energy. We can equally well consider the tensor in ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) or ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) form. We start by considering the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) version of the tensor
(12.2) T ( x ) = T μ ν ( x ) ω μ ω ν (12.2) T ( x ) = T μ ν ( x ) ω μ ω ν {:(12.2)T(x)=T_(mu nu)(x)omega^(mu)oxomega^(nu):}\begin{equation*} \boldsymbol{T}(x)=T_{\mu \nu}(x) \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu} \tag{12.2} \end{equation*}(12.2)T(x)=Tμν(x)ωμων
where, since the tensor is symmetric, T μ ν ( x ) = T ν μ ( x ) . 2 T μ ν ( x ) = T ν μ ( x ) . 2 T_(mu nu)(x)=T_(nu mu)(x).^(2)T_{\mu \nu}(x)=T_{\nu \mu}(x) .{ }^{2}Tμν(x)=Tνμ(x).2
Take the 4 -velocity of an observer to be u u u\boldsymbol{u}u. There is a recipe for extracting physical information by filling in slots of the tensor T ( T ( T(\boldsymbol{T}(T(, ) . ) . ).) .). Start by inserting the velocity into one slot (it doesn't matter which, since the tensor is symmetric), to obtain a 1 -form
(12.4) T ( , u ) = ( 4-momentum density in the observer's rest frame ) . (12.4) T ( , u ) = (  4-momentum density in   the observer's rest frame  ) . {:(12.4)T(","u)=-((" 4-momentum density in ")/(" the observer's rest frame ")).:}\begin{equation*} \boldsymbol{T}(, \boldsymbol{u})=-\binom{\text { 4-momentum density in }}{\text { the observer's rest frame }} . \tag{12.4} \end{equation*}(12.4)T(,u)=( 4-momentum density in  the observer's rest frame ).
Now insert a unit displacement vector a a a\boldsymbol{a}a into the second slot. Since the slots are full, the output is a number. In this case, we obtain
(12.5) T ( a , u ) = ( 4-momentum density component directed along a in the observer's rest frame ) (12.5) T ( a , u ) = (  4-momentum density component directed   along  a  in the observer's rest frame  ) {:(12.5)T(a","u)=-((" 4-momentum density component directed ")/(" along "a" in the observer's rest frame ")):}\begin{equation*} \boldsymbol{T}(\boldsymbol{a}, \boldsymbol{u})=-\binom{\text { 4-momentum density component directed }}{\text { along } \boldsymbol{a} \text { in the observer's rest frame }} \tag{12.5} \end{equation*}(12.5)T(a,u)=( 4-momentum density component directed  along a in the observer's rest frame )
12.1 Another look at the energymomentum tensor 131
12.2 Example energy-momentum tensors
12.3 Classical particles 134
12.4 Conservation laws 136
Chapter summary 139
Exercises 140
1 1 ^(1){ }^{1}1 We expect the Einstein field equation
to read (in words)
Curvature of = = === Energy density
spacetime = = === of matter fields
(12.1)
We will (finally!) justify this in the next chapter, using the tools we have developed.
2 2 ^(2){ }^{2}2 We saw in Chapter 4, how the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor T T T\boldsymbol{T}T could be understood by describing it as an outer product of 1 forms
(12.3) T ( , ) = p ~ ( ) J ~ ( ) (12.3) T ( , ) = p ~ ( ) J ~ ( ) {:(12.3)T(",")= tilde(p)()ox tilde(J)():}\begin{equation*} \boldsymbol{T}(,)=\tilde{\boldsymbol{p}}() \otimes \tilde{\boldsymbol{J}}() \tag{12.3} \end{equation*}(12.3)T(,)=p~()J~()
where p ~ p ~ tilde(p)\tilde{\boldsymbol{p}}p~ is a momentum 1 -form and J ~ J ~ tilde(J)\tilde{J}J~ is a number current 1 -form for particles. We previously took J ~ J ~ tilde(J)\tilde{\boldsymbol{J}}J~ to describe a swarm of non-interacting particles. Although the tensor T T T\boldsymbol{T}T describes energy-momentum more generally than for the case of swarms of particles, this description is useful in understanding T T T\boldsymbol{T}T. For example, if an observer has velocity u u u\boldsymbol{u}u, then we can interpret T ( , u ) = p ~ J ( u ) = p ~ n T ( , u ) = p ~ J ( u ) = p ~ n T(,u)= tilde(p)ox J(u)=- tilde(p)n\boldsymbol{T}(, \boldsymbol{u})=\tilde{\boldsymbol{p}} \otimes \boldsymbol{J}(\boldsymbol{u})=-\tilde{\boldsymbol{p}} nT(,u)=p~J(u)=p~n as (minus) the momentum density of particles in the observer's frame, We would inter he T ( u ) = , p ~ ( u ) J ~ = J ~ T ( u ) = , p ~ ( u ) J ~ = J ~ T(u)=, tilde(p)(u)ox tilde(J)=- tilde(J)\boldsymbol{T}(\boldsymbol{u})=,\tilde{p}(\boldsymbol{u}) \otimes \tilde{J}=-\tilde{J}T(u)=,p~(u)J~=J~ pret T ( u ) = , p ( u ) J = E J T ( u ) = , p ( u ) J = E J T(u)=,p(u)ox J=-EJ\boldsymbol{T}(\boldsymbol{u})=,\boldsymbol{p}(\boldsymbol{u}) \otimes \boldsymbol{J}=-E \boldsymbol{J}T(u)=,p(u)J=EJ as ( mi ( mi (mi(\mathrm{mi}(mi nus) the energy flux. Since the tensor is symmetric, these are the same thing.
3 3 ^(3){ }^{3}3 Since we use the local basis vectors in the observer's orthonormal frame, denoted with hats on the indices, the interpretations here will always apply to measurements made in this observer's measurements made in this observer's frame. We can turn these into up components using the Minkowski tensor η μ ^ ν ^ η μ ^ ν ^ eta^( hat(mu) hat(nu))\eta^{\hat{\mu} \hat{\nu}}ημ^ν^, so we pick up a factor of -1 for each of the two timelike components we raise. This gives us
T 0 ^ 0 ^ = T 0 ^ 0 ^ , T 0 ^ i ^ = T 0 ^ i ^ , (12.7) T i ^ j ^ = T i ^ j ^ . T 0 ^ 0 ^ = T 0 ^ 0 ^ , T 0 ^ i ^ = T 0 ^ i ^ , (12.7) T i ^ j ^ = T i ^ j ^ . {:[T^( hat(0) hat(0))=T_( hat(0) hat(0))","],[T^( hat(0) hat(i))=-T_( hat(0) hat(i))","],[(12.7)T^( hat(i) hat(j))=T_( hat(i) hat(j)).]:}\begin{align*} & T^{\hat{0} \hat{0}}=T_{\hat{0} \hat{0}}, \\ & T^{\hat{0} \hat{i}}=-T_{\hat{0} \hat{i}}, \\ & T^{\hat{i} \hat{j}}=T_{\hat{i} \hat{j}} . \tag{12.7} \end{align*}T0^0^=T0^0^,T0^i^=T0^i^,(12.7)Ti^j^=Ti^j^.
Fig. 12.1 Flux of the i ^ i ^ hat(i)\hat{i}i^ th component of momentum across a surface parallel to ω j ω j omega^(j)\boldsymbol{\omega}^{j}ωj (that is, a surface with normal e j e j e_(j)\boldsymbol{e}_{j}ej ).
Filling, instead, both slots with u u u\boldsymbol{u}u, we obtain
(12.6) T ( u , u ) = ( Energy density in the observer's rest frame ) (12.6) T ( u , u ) = (  Energy density   in the observer's rest frame  ) {:(12.6)T(u","u)=((" Energy density ")/(" in the observer's rest frame ")):}\begin{equation*} \boldsymbol{T}(\boldsymbol{u}, \boldsymbol{u})=\binom{\text { Energy density }}{\text { in the observer's rest frame }} \tag{12.6} \end{equation*}(12.6)T(u,u)=( Energy density  in the observer's rest frame )

Example 12.1

We shall also discuss components T μ ν = T ( e μ , e ν ) T μ ν = T e μ , e ν T_(mu nu)=T(e_(mu),e_(nu))T_{\mu \nu}=\boldsymbol{T}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)Tμν=T(eμ,eν). Timelike components can be extracted by noting that in the observer's orthonormal frame u = e 0 ^ u = e 0 ^ u=e_( hat(0))\boldsymbol{u}=\boldsymbol{e}_{\hat{0}}u=e0^ and so, and perhaps most importantly, 3 3 ^(3){ }^{3}3
(12.8) T ( e 0 ^ , e 0 ^ ) = T 0 ^ 0 ^ = T 0 ^ 0 ^ = ( energy density ) . (12.8) T e 0 ^ , e 0 ^ = T 0 ^ 0 ^ = T 0 ^ 0 ^ = (  energy density  ) . {:(12.8)T(e_( hat(0)),e_( hat(0)))=T_( hat(0) hat(0))=T^( hat(0) hat(0))=(" energy density ").:}\begin{equation*} \boldsymbol{T}\left(\boldsymbol{e}_{\hat{0}}, \boldsymbol{e}_{\hat{0}}\right)=T_{\hat{0} \hat{0}}=T^{\hat{0} \hat{0}}=(\text { energy density }) . \tag{12.8} \end{equation*}(12.8)T(e0^,e0^)=T0^0^=T0^0^=( energy density ).
We also have
(12.9) T ( e i ^ , e 0 ^ ) = T i ^ 0 ^ = T i 0 ^ 0 ^ = ( i th component of 4 -momentum density ) . (12.9) T e i ^ , e 0 ^ = T i ^ 0 ^ = T i 0 ^ 0 ^ = ( i  th component of  4 -momentum density  ) . {:(12.9)T(e_( hat(i)),e_( hat(0)))=T_( hat(i) hat(0))=-T^(i hat(0) hat(0))=-((i" th component of ")/(4"-momentum density ")).:}\begin{equation*} \boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{0}}\right)=T_{\hat{i} \hat{0}}=-T^{i \hat{0} \hat{0}}=-\binom{i \text { th component of }}{4 \text {-momentum density }} . \tag{12.9} \end{equation*}(12.9)T(ei^,e0^)=Ti^0^=Ti0^0^=(i th component of 4-momentum density ).
which is also equivalent to an energy flux in the i i iii th direction. We can also interpret the purely spatial components.
(12.10) T ( e i ^ , e j ^ ) = T i ^ j ^ = T i ^ j ^ = ( i th component of 4-momentum flux crossing a surface parallel to ω j ^ ) . (12.10) T e i ^ , e j ^ = T i ^ j ^ = T i ^ j ^ = ( i  th component of 4-momentum flux   crossing a surface parallel to  ω j ^ ) . {:(12.10)T(e_( hat(i)),e_( hat(j)))=T_( hat(i) hat(j))=T^( hat(i) hat(j))=((i" th component of 4-momentum flux ")/(" crossing a surface parallel to "omega^( hat(j)))).:}\begin{equation*} \boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{j}}\right)=T_{\hat{i} \hat{j}}=T^{\hat{i} \hat{j}}=\binom{i \text { th component of 4-momentum flux }}{\text { crossing a surface parallel to } \boldsymbol{\omega}^{\hat{j}}} . \tag{12.10} \end{equation*}(12.10)T(ei^,ej^)=Ti^j^=Ti^j^=(i th component of 4-momentum flux  crossing a surface parallel to ωj^).
The idea of the latter is shown in Fig. 12.1. From the version of the energy-momentum tensor in the sidenote, we have that T i ^ j ^ = p ~ ( e i ^ ) J ~ ( e j ~ ) = p i ^ J j ^ T i ^ j ^ = p ~ e i ^ J ~ e j ~ = p i ^ J j ^ T_( hat(i) hat(j))= tilde(p)(e_( hat(i))) tilde(J)(e_( tilde(j)))=p_( hat(i))J_( hat(j))T_{\hat{i} \hat{j}}=\tilde{p}\left(e_{\hat{i}}\right) \tilde{J}\left(e_{\tilde{j}}\right)=p_{\hat{i}} J_{\hat{j}}Ti^j^=p~(ei^)J~(ej~)=pi^Jj^, which is consistent with this interpretation. However, another interpretation is possible. Since rate of momentum is equivalent to force, we can also write
(12.11) T ( e i ^ , e j ^ ) = T i ^ j ^ = T i ^ j ^ = ( i th component of force across a surface parallel to ω j ^ ) . (12.11) T e i ^ , e j ^ = T i ^ j ^ = T i ^ j ^ = ( i  th component of force   across a surface parallel to  ω j ^ ) . {:(12.11)T(e_( hat(i)),e_( hat(j)))=T_( hat(i) hat(j))=T^( hat(i) hat(j))=((i" th component of force ")/(" across a surface parallel to "omega^( hat(j)))).:}\begin{equation*} \boldsymbol{T}\left(\boldsymbol{e}_{\hat{i}}, \boldsymbol{e}_{\hat{j}}\right)=T_{\hat{i} \hat{j}}=T^{\hat{i} \hat{j}}=\binom{i \text { th component of force }}{\text { across a surface parallel to } \boldsymbol{\omega}^{\hat{j}}} . \tag{12.11} \end{equation*}(12.11)T(ei^,ej^)=Ti^j^=Ti^j^=(i th component of force  across a surface parallel to ωj^).
A cartoon of the components of the energy-momentum tensor is shown in Fig. 12.2.
Fig. 12.2 The components of the energy-momentum tensor in the orthonormal frame of an observer.

12.2 Example energy-momentum tensors

The general discussion of the energy-momentum tensor is all very well, but of limited use without some idea of what the various components of T T T\boldsymbol{T}T actually are. We now examine some examples of how to build 4 T 4 T ^(4)T{ }^{4} \boldsymbol{T}4T. Since gravity works over long length scales where the granularity of matter is often not too important, we will mostly be concerned with
continuous fields of matter and so we need the energy-momentum tensor for this purpose. The continuous version of the particle matter we have considered so far is a fluid. We will consider a so-called perfect fluid, which is continuous matter with energy density ρ ρ rho\rhoρ and an isotropic pressure 5 p 5 p ^(5)p{ }^{5} p5p in its rest frame, but no other interactions. 6 6 ^(6){ }^{6}6

Example 12.2

First, consider the case of the fluid at rest in a local inertial reference frame, which has metric η η eta\boldsymbol{\eta}η. We have an energy density T 0 ^ 0 ^ = T 0 ^ 0 ^ = ρ T 0 ^ 0 ^ = T 0 ^ 0 ^ = ρ T^( hat(0) hat(0))=T_( hat(0) hat(0))=rhoT^{\hat{0} \hat{0}}=T_{\hat{0} \hat{0}}=\rhoT0^0^=T0^0^=ρ, where ρ ρ rho\rhoρ is the total energy-density of the fluid, as measured in the fluid's rest frame. The force in the i ^ i ^ hat(i)\hat{i}i^ th direction, perpendicular to the surface whose normal is in the i ^ i ^ hat(i)\hat{i}i^ direction, is simply the definition of the pressure p p ppp. We therefore have T i i = T i ^ i ^ = p T i i = T i ^ i ^ = p T^(ii)=T_( hat(i) hat(i))=pT^{i i}=T_{\hat{i} \hat{i}}=pTii=Ti^i^=p and can write
(12.12) T μ ^ ν ^ = ( ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p ) (12.12) T μ ^ ν ^ = ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p {:(12.12)T_( hat(mu) hat(nu))=([rho,0,0,0],[0,p,0,0],[0,0,p,0],[0,0,0,p]):}T_{\hat{\mu} \hat{\nu}}=\left(\begin{array}{cccc} \rho & 0 & 0 & 0 \tag{12.12}\\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array}\right)(12.12)Tμ^ν^=(ρ0000p0000p0000p)
The velocity 4 -vector u u u\boldsymbol{u}u of the fluid fixes the timelike basis vector according to e 0 ^ = e 0 ^ = e_( hat(0))=\boldsymbol{e}_{\hat{0}}=e0^=
u u u\boldsymbol{u}u. In the fluid's rest frame, the velocity 1 -form u ~ = u μ ^ ω μ u ~ = u μ ^ ω μ tilde(u)=u_( hat(mu))omega^(mu)\tilde{\boldsymbol{u}}=u_{\hat{\mu}} \boldsymbol{\omega}^{\mu}u~=uμ^ωμ has components u μ ^ = u μ ^ = u_( hat(mu))=u_{\hat{\mu}}=uμ^= ( 1 , 0 , 0 , 0 ) ( 1 , 0 , 0 , 0 ) (-1,0,0,0)(-1,0,0,0)(1,0,0,0). The ( 0,2 ) tensor product u ~ u ~ u ~ u ~ tilde(u)ox tilde(u)\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}}u~u~ has components
(12.13) ( u ~ u ~ ) μ ^ ν ^ = u ~ ( e μ ^ ) u ~ ( e ν ^ ) = u μ ^ u ν ^ (12.13) ( u ~ u ~ ) μ ^ ν ^ = u ~ e μ ^ u ~ e ν ^ = u μ ^ u ν ^ {:(12.13)( tilde(u)ox tilde(u))_( hat(mu) hat(nu))= tilde(u)(e_( hat(mu))) tilde(u)(e_( hat(nu)))=u_( hat(mu))u_( hat(nu)):}\begin{equation*} (\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}})_{\hat{\mu} \hat{\nu}}=\tilde{\boldsymbol{u}}\left(\boldsymbol{e}_{\hat{\mu}}\right) \tilde{\boldsymbol{u}}\left(\boldsymbol{e}_{\hat{\nu}}\right)=u_{\hat{\mu}} u_{\hat{\nu}} \tag{12.13} \end{equation*}(12.13)(u~u~)μ^ν^=u~(eμ^)u~(eν^)=uμ^uν^
and so we have
(12.14) ( u ~ u ~ ) μ ^ ν ^ = ( 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) (12.14) ( u ~ u ~ ) μ ^ ν ^ = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 {:(12.14)( tilde(u)ox tilde(u))_( hat(mu) hat(nu))=([1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]):}(\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}})_{\hat{\mu} \hat{\nu}}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \tag{12.14}\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)(12.14)(u~u~)μ^ν^=(1000000000000000)
We therefore find we can recreate the energy-momentum matrix via the decomposition
T μ ^ ν ^ = ( ρ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) + ( 0 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p ) (12.15) = ρ ( u ~ u ~ ) μ ^ ν ^ + p [ η μ ^ ν ^ + ( u ~ u ~ ) μ ^ ν ^ ] . T μ ^ ν ^ = ρ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p (12.15) = ρ ( u ~ u ~ ) μ ^ ν ^ + p η μ ^ ν ^ + ( u ~ u ~ ) μ ^ ν ^ . {:[T_( hat(mu) hat(nu))=([rho,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0])+([0,0,0,0],[0,p,0,0],[0,0,p,0],[0,0,0,p])],[(12.15)=rho( tilde(u)ox tilde(u))_( hat(mu) hat(nu))+p[eta_( hat(mu) hat(nu))+(( tilde(u))ox( tilde(u)))_( hat(mu) hat(nu))].]:}\begin{align*} T_{\hat{\mu} \hat{\nu}} & =\left(\begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)+\left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array}\right) \\ & =\rho(\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}})_{\hat{\mu} \hat{\nu}}+p\left[\eta_{\hat{\mu} \hat{\nu}}+(\tilde{\boldsymbol{u}} \otimes \tilde{\boldsymbol{u}})_{\hat{\mu} \hat{\nu}}\right] . \tag{12.15} \end{align*}Tμ^ν^=(ρ000000000000000)+(00000p0000p0000p)(12.15)=ρ(u~u~)μ^ν^+p[ημ^ν^+(u~u~)μ^ν^].
The components of this equation balance 7 7 ^(7){ }^{7}7 and so it represents a valid tensor equation.
We conclude that for a perfect fluid in a local inertial frame (or in flat spacetime) we have a equation for the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor T T T\boldsymbol{T}T, which we can write as
(12.16) T ( , ) = ( ρ + p ) u ~ ( ) u ~ ( ) + p η ( , ) (12.16) T ( , ) = ( ρ + p ) u ~ ( ) u ~ ( ) + p η ( , ) {:(12.16)T(",")=(rho+p) tilde(u)()ox tilde(u)()+p eta(","):}\begin{equation*} \boldsymbol{T}(,)=(\rho+p) \tilde{\boldsymbol{u}}() \otimes \tilde{\boldsymbol{u}}()+p \boldsymbol{\eta}(,) \tag{12.16} \end{equation*}(12.16)T(,)=(ρ+p)u~()u~()+pη(,)
where we take u ~ u ~ tilde(u)\tilde{\boldsymbol{u}}u~ to be the velocity 1 -form of the fluid itself. The tensor T T T\boldsymbol{T}T inputs two vectors to generate a number. It has components
(12.17) T μ ^ ν ^ = ( ρ + p ) u μ ^ u ν ^ + p η μ ^ ν ^ (12.17) T μ ^ ν ^ = ( ρ + p ) u μ ^ u ν ^ + p η μ ^ ν ^ {:(12.17)T_( hat(mu) hat(nu))=(rho+p)u_( hat(mu))u_( hat(nu))+peta_( hat(mu) hat(nu)):}\begin{equation*} T_{\hat{\mu} \hat{\nu}}=(\rho+p) u_{\hat{\mu}} u_{\hat{\nu}}+p \eta_{\hat{\mu} \hat{\nu}} \tag{12.17} \end{equation*}(12.17)Tμ^ν^=(ρ+p)uμ^uν^+pημ^ν^
Example 12.3
Taking the trace T T TTT of the tensor gives us a number
(12.18) T = T ν ^ ν ^ = T μ ^ ν ^ η μ ^ ν ^ = ( ρ + p ) + 4 p = ρ + 3 p (12.18) T = T ν ^ ν ^ = T μ ^ ν ^ η μ ^ ν ^ = ( ρ + p ) + 4 p = ρ + 3 p {:(12.18)T=T_( hat(nu))^( hat(nu))=T_( hat(mu) hat(nu))eta^( hat(mu) hat(nu))=-(rho+p)+4p=-rho+3p:}\begin{equation*} T=T_{\hat{\nu}}^{\hat{\nu}}=T_{\hat{\mu} \hat{\nu}} \eta^{\hat{\mu} \hat{\nu}}=-(\rho+p)+4 p=-\rho+3 p \tag{12.18} \end{equation*}(12.18)T=Tν^ν^=Tμ^ν^ημ^ν^=(ρ+p)+4p=ρ+3p
Thermodynamics says that a photon gas has the equation of state ρ = 3 p ρ = 3 p rho=3p\rho=3 pρ=3p, so we conclude that the energy-momentum tensor for the photon gas is traceless ( T = 0 ) . 8 ( T = 0 ) . 8 (T=0).^(8)(T=0) .{ }^{8}(T=0).8
5 5 ^(5){ }^{5}5 Do not confuse pressure p p ppp with momentum, which regrettably has the same symbol.
6 6 ^(6){ }^{6}6 The total energy density ρ ρ rho\rhoρ will generally include a contribution from the mass density ρ 0 ρ 0 rho_(0)\rho_{0}ρ0 of the fluid, along with a contribution from the internal energy density. The latter results from internal motion and interactions between the constituents of the fluid. If there is no internal energy then pressure p = 0 p = 0 p=0p=0p=0 and we call the matter dust, which has ρ = ρ 0 ρ = ρ 0 rho=rho_(0)\rho=\rho_{0}ρ=ρ0. In contrast, a perfect fluid has an isotropic pressure p p ppp in its rest frame, but no viscosity or shear stresses (and now ρ ρ 0 ρ ρ 0 rho!=rho_(0)\rho \neq \rho_{0}ρρ0 ). The relationship between ρ ρ rho\rhoρ and p p ppp for the fluid is known as its equation of state.
7 7 ^(7){ }^{7}7 That is, we have μ ^ ν ^ μ ^ ν ^ hat(mu) hat(nu)\hat{\mu} \hat{\nu}μ^ν^ in the down position on both sides. Recall that we say such equations are manifestly covariant and are therefore 'valid tensor equations'.
8 8 ^(8){ }^{8}8 A field with an energy-momentum tensor whose trace vanishes is expected to have massless excitations. This is true for the electromagnetic field, whose excitations, photons, are certainly massless. The vanishing trace can be linked to scale invariance, as discussed in the book by Zee.
The equivalence principle tells us that a valid tensor equation in a flat space is a valid tensor equation in a curved space as long as we replace the Minkowski metric tensor η η eta\boldsymbol{\eta}η with the generalized metric tensor g g g\boldsymbol{g}g. We can then upgrade eqn 12.16 to the generalized version in curved space, where it becomes
T ( , ) = ( ρ + p ) u ~ ( ) u ~ ( ) + p g ( , ) , (12.19) T μ ν = ( ρ + p ) u μ u ν + p g μ ν . T ( , ) = ( ρ + p ) u ~ ( ) u ~ ( ) + p g ( , ) , (12.19) T μ ν = ( ρ + p ) u μ u ν + p g μ ν . {:[T(",")=(rho+p) tilde(u)()ox tilde(u)()+pg(",")","],[(12.19)T_(mu nu)=(rho+p)u_(mu)u_(nu)+pg_(mu nu).]:}\begin{align*} \boldsymbol{T}(,) & =(\rho+p) \tilde{\boldsymbol{u}}() \otimes \tilde{\boldsymbol{u}}()+p \boldsymbol{g}(,), \\ T_{\mu \nu} & =(\rho+p) u_{\mu} u_{\nu}+p g_{\mu \nu} . \tag{12.19} \end{align*}T(,)=(ρ+p)u~()u~()+pg(,),(12.19)Tμν=(ρ+p)uμuν+pgμν.
Since tensor equations in physics are generally valid with respect to transforming all indices from up to down, and vice versa, we shall also use the ( 2,0 ) version of T T T\boldsymbol{T}T whose components in a general coordinate frame are T μ ν = ( ρ + p ) u μ u ν + p g μ ν T μ ν = ( ρ + p ) u μ u ν + p g μ ν T^(mu nu)=(rho+p)u^(mu)u^(nu)+pg^(mu nu)T^{\mu \nu}=(\rho+p) u^{\mu} u^{\nu}+p g^{\mu \nu}Tμν=(ρ+p)uμuν+pgμν. In comparing quantities with up and down indices, remember that, as usual, indices are raised and lowered using the components of the metric tensor.

12.3 Classical particles

Einstein's equation is expressed in terms of classical fields. We shall find that the matter field corresponding to a perfect fluid is the most useful for our applications. However, in addition to fluids, we can also treat the motion of point-like particles. The most simple example is introduced below.

Example 12.4

Consider a swarm of n n nnn (non-interacting) dust particles per unit volume in flat space, each of rest mass m m mmm, moving with the same 4 -velocity with components ( γ , γ v x , γ v y , 0 ) γ , γ v x , γ v y , 0 (gamma,gammav^(x),gammav^(y),0)\left(\gamma, \gamma v^{x}, \gamma v^{y}, 0\right)(γ,γvx,γvy,0). Temporarily restoring factors of c c ccc, we have total energy density ( γ , γ v x , γ v y , 0 ) γ , γ v x , γ v y , 0 (gamma,gammav^(x),gammav^(y),0)\left(\gamma, \gamma v^{x}, \gamma v^{y}, 0\right)(γ,γvx,γvy,0). Temporarily restoring factors of c c ccc, we have total energy density
T 00 = n m γ c 2 T 00 = n m γ c 2 T^(00)=nm gammac^(2)T^{00}=n m \gamma c^{2}T00=nmγc2. The momentum density in the i i iii-direction is T 0 i = T i 0 = n m γ c v i T 0 i = T i 0 = n m γ c v i T^(0i)=T^(i0)=nm gamma cv^(i)T^{0 i}=T^{i 0}=n m \gamma c v^{i}T0i=Ti0=nmγcvi.
The momentum flux in, say, the x x xxx-direction, crossing the surface parallel to ω y ω y omega^(y)\boldsymbol{\omega}^{y}ωy is
T x y = n m γ v x v y T x y = n m γ v x v y T^(xy)=nm gammav^(x)v^(y)T^{x y}=n m \gamma v^{x} v^{y}Txy=nmγvxvy. This is enough for us to construct the tensor, whose components are
(12.20) T μ ν = n m γ ( c 2 v x c v y c 0 v x c ( v x ) 2 v x v y 0 v y c v x v y ( v y ) 2 0 0 0 0 0 ) (12.20) T μ ν = n m γ c 2 v x c v y c 0 v x c v x 2 v x v y 0 v y c v x v y v y 2 0 0 0 0 0 {:(12.20)T^(mu nu)=nm gamma([c^(2),v^(x)c,v^(y)c,0],[v^(x)c,(v^(x))^(2),v^(x)v^(y),0],[v^(y)c,v^(x)v^(y),(v^(y))^(2),0],[0,0,0,0]):}T^{\mu \nu}=n m \gamma\left(\begin{array}{cccc} c^{2} & v^{x} c & v^{y} c & 0 \tag{12.20}\\ v^{x} c & \left(v^{x}\right)^{2} & v^{x} v^{y} & 0 \\ v^{y} c & v^{x} v^{y} & \left(v^{y}\right)^{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)(12.20)Tμν=nmγ(c2vxcvyc0vxc(vx)2vxvy0vycvxvy(vy)200000)
In order to treat particles more generally using the techniques of field theory, we can use the fact that particles appear on a world line z μ ( τ ) z μ ( τ ) z^(mu)(tau)z^{\mu}(\tau)zμ(τ), where τ τ tau\tauτ is the proper time parametrizing the world line. 9 9 ^(9){ }^{9}9 In order to select that part of spacetime that the particle intersects, we use a delta function to pick out the curve z μ ( τ ) z μ ( τ ) z^(mu)(tau)z^{\mu}(\tau)zμ(τ). The particles are assumed noninteracting here, so the only contribution to their energy is their mass. To capture the mass density field ρ 0 ( x ) ρ 0 ( x ) rho_(0)(x)\rho_{0}(x)ρ0(x) of a particle of mass m m mmm in flat spacetime we write
(12.21) ρ 0 ( x ) = m d τ δ ( 4 ) [ x z ( τ ) ] (12.21) ρ 0 ( x ) = m d τ δ ( 4 ) [ x z ( τ ) ] {:(12.21)rho_(0)(x)=m intdtaudelta^((4))[x-z(tau)]:}\begin{equation*} \rho_{0}(x)=m \int \mathrm{~d} \tau \delta^{(4)}[x-z(\tau)] \tag{12.21} \end{equation*}(12.21)ρ0(x)=m dτδ(4)[xz(τ)]
where, in Cartesian coordinates, we have defined a four-dimensional delta-function
δ ( 4 ) [ x z ( τ ) ] = δ [ x 0 z 0 ( τ ) ] δ [ x 1 z 1 ( τ ) ] δ [ x 2 z 2 ( τ ) ] δ [ x 3 z 3 ( τ ) ] . δ ( 4 ) [ x z ( τ ) ] = δ x 0 z 0 ( τ ) δ x 1 z 1 ( τ ) δ x 2 z 2 ( τ ) δ x 3 z 3 ( τ ) . delta^((4))[x-z(tau)]=delta[x^(0)-z^(0)(tau)]delta[x^(1)-z^(1)(tau)]delta[x^(2)-z^(2)(tau)]delta[x^(3)-z^(3)(tau)].\delta^{(4)}[x-z(\tau)]=\delta\left[x^{0}-z^{0}(\tau)\right] \delta\left[x^{1}-z^{1}(\tau)\right] \delta\left[x^{2}-z^{2}(\tau)\right] \delta\left[x^{3}-z^{3}(\tau)\right] .δ(4)[xz(τ)]=δ[x0z0(τ)]δ[x1z1(τ)]δ[x2z2(τ)]δ[x3z3(τ)].
( 12.22 ) ( 12.22 ) (12.22)(12.22)(12.22)

Example 12.5

To see how the definition works, we change the integration variable from τ τ tau\tauτ to z 0 ( τ ) z 0 ( τ ) z^(0)(tau)z^{0}(\tau)z0(τ) using u 0 = d z 0 / d τ u 0 = d z 0 / d τ u^(0)=dz^(0)//dtauu^{0}=\mathrm{d} z^{0} / \mathrm{d} \tauu0=dz0/dτ to write
ρ 0 ( x ) = m d z 0 δ ( 4 ) [ x z ( z 0 ( τ ) ) ] u 0 = m d z 0 δ ( 3 ) [ x z ( z 0 ( τ ) ) ] u 0 δ [ t z 0 ( τ ) ] (12.23) = m u 0 δ ( 3 ) [ x z ( τ ) ] , ρ 0 ( x ) = m d z 0 δ ( 4 ) x z z 0 ( τ ) u 0 = m d z 0 δ ( 3 ) x z z 0 ( τ ) u 0 δ t z 0 ( τ ) (12.23) = m u 0 δ ( 3 ) [ x z ( τ ) ] , {:[rho_(0)(x)=m intdz^(0)(delta^((4))[x-z(z^(0)(tau))])/(u^(0))],[=m intdz^(0)(delta^((3))[x-z(z^(0)(tau))])/(u^(0))delta[t-z^(0)(tau)]],[(12.23)=(m)/(u^(0))delta^((3))[x-z(tau)]","]:}\begin{align*} \rho_{0}(x) & =m \int \mathrm{~d} z^{0} \frac{\delta^{(4)}\left[x-z\left(z^{0}(\tau)\right)\right]}{u^{0}} \\ & =m \int \mathrm{~d} z^{0} \frac{\delta^{(3)}\left[x-z\left(z^{0}(\tau)\right)\right]}{u^{0}} \delta\left[t-z^{0}(\tau)\right] \\ & =\frac{m}{u^{0}} \delta^{(3)}[x-z(\tau)], \tag{12.23} \end{align*}ρ0(x)=m dz0δ(4)[xz(z0(τ))]u0=m dz0δ(3)[xz(z0(τ))]u0δ[tz0(τ)](12.23)=mu0δ(3)[xz(τ)],
where τ τ tau\tauτ in the final line solves the equation z 0 ( τ ) = t z 0 ( τ ) = t z^(0)(tau)=tz^{0}(\tau)=tz0(τ)=t. This is now simply a function that finds the particle in three-dimensional space. Notice how the factor u 0 u 0 u^(0)u^{0}u0, which in flat space is simply γ γ gamma\gammaγ, deals with the length contraction of the three-dimensional volume that determines the value of the mass density (see Fig. 2.7).
We can also write the mass current, which is most generally written as a field J m ( x ) = ρ 0 ( x ) u ( x ) J m ( x ) = ρ 0 ( x ) u ( x ) J_(m)(x)=rho_(0)(x)u(x)\boldsymbol{J}_{m}(x)=\rho_{0}(x) \boldsymbol{u}(x)Jm(x)=ρ0(x)u(x), where u ( x ) u ( x ) u(x)\boldsymbol{u}(x)u(x) is the velocity of the matter at x x xxx. In terms of particles, we have the component expression 10 10 ^(10){ }^{10}10
(12.25) J m α ( x ) = m d τ u α δ ( 4 ) [ x z ( τ ) ] (12.25) J m α ( x ) = m d τ u α δ ( 4 ) [ x z ( τ ) ] {:(12.25)J_(m)^(alpha)(x)=m intdtauu^(alpha)delta^((4))[x-z(tau)]:}\begin{equation*} J_{m}^{\alpha}(x)=m \int \mathrm{~d} \tau u^{\alpha} \delta^{(4)}[x-z(\tau)] \tag{12.25} \end{equation*}(12.25)Jmα(x)=m dτuαδ(4)[xz(τ)]
In the same way, we write a flat-space energy-momentum tensor with components
(12.26) T α β ( x ) = m d τ u α u β δ ( 4 ) [ x z ( τ ) ] (12.26) T α β ( x ) = m d τ u α u β δ ( 4 ) [ x z ( τ ) ] {:(12.26)T^(alpha beta)(x)=m intdtauu^(alpha)u^(beta)delta^((4))[x-z(tau)]:}\begin{equation*} T^{\alpha \beta}(x)=m \int \mathrm{~d} \tau u^{\alpha} u^{\beta} \delta^{(4)}[x-z(\tau)] \tag{12.26} \end{equation*}(12.26)Tαβ(x)=m dτuαuβδ(4)[xz(τ)]
which is simply a version of eqn 12.19 for particles with p = 0 p = 0 p=0p=0p=0.

Example 12.6

In curved spacetime, we must recall that g d 4 x g d 4 x sqrt(-g)d^(4)x\sqrt{-g} \mathrm{~d}^{4} xg d4x is the invariant volume element, so δ ( 4 ) ( x y ) / g δ ( 4 ) ( x y ) / g delta^((4))(x-y)//sqrt(-g)\delta^{(4)}(x-y) / \sqrt{-g}δ(4)(xy)/g is a scalar. The expressions above become, in curved spacetime,
ρ 0 ( x ) = m d τ δ ( 4 ) [ x z ( τ ) ] g J m α ( x ) = m d τ u α δ ( 4 ) [ x z ( τ ) ] g (12.27) T α β ( x ) = m d τ u α u β δ ( 4 ) [ x z ( τ ) ] g ρ 0 ( x ) = m d τ δ ( 4 ) [ x z ( τ ) ] g J m α ( x ) = m d τ u α δ ( 4 ) [ x z ( τ ) ] g (12.27) T α β ( x ) = m d τ u α u β δ ( 4 ) [ x z ( τ ) ] g {:[rho_(0)(x)=m intdtau(delta^((4))[x-z(tau)])/(sqrt(-g))],[J_(m)^(alpha)(x)=m intdtauu^(alpha)(delta^((4))[x-z(tau)])/(sqrt(-g))],[(12.27)T^(alpha beta)(x)=m intdtauu^(alpha)u^(beta)(delta^((4))[x-z(tau)])/(sqrt(-g))]:}\begin{align*} \rho_{0}(x) & =m \int \mathrm{~d} \tau \frac{\delta^{(4)}[x-z(\tau)]}{\sqrt{-g}} \\ J_{m}^{\alpha}(x) & =m \int \mathrm{~d} \tau u^{\alpha} \frac{\delta^{(4)}[x-z(\tau)]}{\sqrt{-g}} \\ T^{\alpha \beta}(x) & =m \int \mathrm{~d} \tau u^{\alpha} u^{\beta} \frac{\delta^{(4)}[x-z(\tau)]}{\sqrt{-g}} \tag{12.27} \end{align*}ρ0(x)=m dτδ(4)[xz(τ)]gJmα(x)=m dτuαδ(4)[xz(τ)]g(12.27)Tαβ(x)=m dτuαuβδ(4)[xz(τ)]g
10 10 ^(10){ }^{10}10 By the same token we can also define a charge current with components in flat space of
(12.24) J q α ( x ) = q d τ u α δ ( 4 ) [ x z ( τ ) ] (12.24) J q α ( x ) = q d τ u α δ ( 4 ) [ x z ( τ ) ] {:(12.24)J_(q)^(alpha)(x)=q intdtauu^(alpha)delta^((4))[x-z(tau)]:}\begin{equation*} J_{q}^{\alpha}(x)=q \int \mathrm{~d} \tau u^{\alpha} \delta^{(4)}[x-z(\tau)] \tag{12.24} \end{equation*}(12.24)Jqα(x)=q dτuαδ(4)[xz(τ)]
where q q qqq is the scalar charge.

12.4 Conservation laws

11 11 ^(11){ }^{11}11 Restoring factors of c c ccc we have the 4 -vector J μ = ( ρ c , J ) J μ = ( ρ c , J ) J^(mu)=(rho c, vec(J))J^{\mu}=(\rho c, \vec{J})Jμ=(ρc,J). The spacelike part J J vec(J)\vec{J}J therefore has units of charge density times velocity.
12 12 ^(12){ }^{12}12 Integrating eqn 12.30 over a flatspace 3 -volume and invoking the divergence theorem gives the conserva tion law in three-dimensional Cartesian space as
(12.29) Q t = S J d S (12.29) Q t = S J d S {:(12.29)-(del Q)/(del t)=int_(S) vec(J)*d vec(S):}\begin{equation*} -\frac{\partial Q}{\partial t}=\int_{S} \vec{J} \cdot \mathrm{~d} \vec{S} \tag{12.29} \end{equation*}(12.29)Qt=SJ dS
where S S SSS is the boundary of the volume V V VVV, whose area element is written as d S d S d vec(S)\mathrm{d} \vec{S}dS In words, the rate of change of charge in the volume is equal to the flux of particles through its surface.
13 13 ^(13){ }^{13}13 The 3-surface could be oriented, so that we have a sense of direction de fined from one side of the surface to the other, just as one has for the 2-surface in three-dimensions.
14 14 ^(14){ }^{14}14 This can be interpreted as saying that all the charge that enters the vol ume in the past part of its boundary (where J J J\boldsymbol{J}J and d Σ d Σ dSigma\mathrm{d} \boldsymbol{\Sigma}dΣ might point in opposite directions) has to exit out of the future part of its boundary (where J J J\boldsymbol{J}J and d Σ d Σ dSigma\mathrm{d} \boldsymbol{\Sigma}dΣ might point in the same direction).
Fig. 12.3 A volume V V V\mathcal{V}V in flat spacetime is bounded by a 3 -surface V V delV\partial \mathcal{V}V which can be described by lots of infinitesimal surface vectors d Σ d Σ dSigma\mathrm{d} \boldsymbol{\Sigma}dΣ. The current density 4 4 4-4-4 vector J J J\boldsymbol{J}J flows into and out of the surface, subject to a global conservation law (eqn 12.32).
Example 12.7
In general, a current vector J J J\boldsymbol{J}J has components J μ = ( ρ , J ) J μ = ( ρ , J ) J^(mu)=(rho, vec(J))J^{\mu}=(\rho, \vec{J})Jμ=(ρ,J). The timelike component ρ ρ rho\rhoρ is called a charge density. In flat space, the integral of this quantity over the 3 -volume V V VVV is the conserved charge
(12.28) Q = V d 3 x ρ (12.28) Q = V d 3 x ρ {:(12.28)Q=int_(V)d^(3)x rho:}\begin{equation*} Q=\int_{V} \mathrm{~d}^{3} x \rho \tag{12.28} \end{equation*}(12.28)Q=V d3xρ
This charge might be electrical charge, or might be the number density of particles. The spacelike components represents the flux of the charge, that is, the number of charges that cross unit area per unit time. 11 11 ^(11){ }^{11}11 Again working in flat spacetime, local conservation implies that J J J\boldsymbol{J}J obeys the equation 12 12 ^(12){ }^{12}12
(12.30) ρ t + J = 0 (12.30) ρ t + J = 0 {:(12.30)(del rho)/(del t)+ vec(grad)* vec(J)=0:}\begin{equation*} \frac{\partial \rho}{\partial t}+\vec{\nabla} \cdot \vec{J}=0 \tag{12.30} \end{equation*}(12.30)ρt+J=0
or,
(12.31) μ J μ = 0 , or equivalently, J μ , μ = 0 (12.31) μ J μ = 0 ,  or equivalently,  J μ , μ = 0 {:(12.31)del_(mu)J^(mu)=0","quad" or equivalently, "quadJ^(mu)_(,mu)=0:}\begin{equation*} \partial_{\mu} J^{\mu}=0, \quad \text { or equivalently, } \quad J^{\mu}{ }_{, \mu}=0 \tag{12.31} \end{equation*}(12.31)μJμ=0, or equivalently, Jμ,μ=0
We shall often write this in vector notation as J = 0 J = 0 grad*J=0\boldsymbol{\nabla} \cdot \boldsymbol{J}=0J=0 (i.e. the divergence of J J J\boldsymbol{J}J is zero). This expression guarantees that the rate of change of charge is equal to the amount of the charge flowing into, or out of, an element of volume. We call this local conservation of charge. We can't therefore have charge disappear on one point and then appear at some other arbitrary point in the Universe, which is what we mean by 'local' in our definition. (Conversely, if we wanted to guarantee conservation of charge, but allow charges to arbitrarily transport across the universe, we would demand only global conservation of a charge.)
We can think about this problem another way. Imagine a flat 3 -surface which lies in the xyt plane, perpendicular to the z z zzz-axis. Then J z J z J^(z)J^{z}Jz represents the total charge that flows across this surface per unit area, per unit time. If this 3 -surface had dimensions Σ z = a × b × τ Σ z = a × b × τ Sigma_(z)=a xx b xx tau\Sigma_{z}=a \times b \times \tauΣz=a×b×τ (i.e. a × b a × b a xx ba \times ba×b in the x y x y xyx yxy plane and τ τ tau\tauτ in the time dimension) then J z Σ z J z Σ z J^(z)Sigma_(z)J^{z} \Sigma_{z}JzΣz would represent the total charge flowing through this 3 -surface (we've multiplied through by Σ z Σ z Sigma_(z)\Sigma_{z}Σz so it is no longer 'per unit area, per unit time'). We could generalize this idea (which has privileged a particular slice in space time) and think of J α Σ α J α Σ α J^(alpha)Sigma_(alpha)J^{\alpha} \Sigma_{\alpha}JαΣα as the total charge that flows across any flat 3 -surface described by a 4 -vector Σ Σ Sigma\boldsymbol{\Sigma}Σ normal to the 3-surface. 13 13 ^(13){ }^{13}13 We could make this even more general by allowing the 3 -surface to be curved, and then the total charge that flows across it could be written as J μ d Σ μ J μ d Σ μ intJ^(mu)dSigma_(mu)\int J^{\mu} \mathrm{d} \Sigma_{\mu}JμdΣμ; here, both J J J\boldsymbol{J}J and Σ Σ Sigma\boldsymbol{\Sigma}Σ are now functions of spacetime.
Now consider some 4 -volume of spacetime V V V\mathcal{V}V, bounded by a closed 3 -surface which we can write as V V delV\partial \mathcal{V}V (see Fig. 12.3). Integrating over the closed surface V V delV\partial \mathcal{V}V, we can then express the global law of charge conservation as 14 14 ^(14){ }^{14}14
(12.32) V J μ d Σ μ = 0 (12.32) V J μ d Σ μ = 0 {:(12.32)int_(delV)J^(mu)dSigma_(mu)=0:}\begin{equation*} \int_{\partial \mathcal{V}} J^{\mu} \mathrm{d} \Sigma_{\mu}=0 \tag{12.32} \end{equation*}(12.32)VJμdΣμ=0
This expresses global charge conservation in flat spacetime, since our 4 -volume V V V\mathcal{V}V can be of any size.
In setting this problem up, we have been thinking about the conservation of electric charge, but this approach would work with any conserved scalar quantity. For example, it could be a number-flux of baryons (which one could write as N = ( N 0 , N ) N = N 0 , N N=(N^(0),( vec(N)))N=\left(N^{0}, \vec{N}\right)N=(N0,N), with N 0 N 0 N_(0)N_{0}N0 the baryon number density and N N NNN the baryon number 3-flux) or the flux of rest mass [ ρ 0 ( 1 , u ) ρ 0 ( 1 , u ) [rho_(0)(1,( vec(u))):}\left[\rho_{0}(1, \vec{u})\right.[ρ0(1,u), with ρ 0 ρ 0 rho_(0)\rho_{0}ρ0 the rest mass density and u u vec(u)\vec{u}u the 3 -velocity]. For both these cases, an equation analogous to eqn 12.32 would follow.
Energy and momentum are conserved, so we might expect a local conservation law for the energy-momentum tensor (which describes the flow of 4 -momentum), just as we found for J J J\boldsymbol{J}J (which describes the flow of
charge). By analogy with the previous example, the local conservation law in flat spacetime can be written in three (equivalent) ways
(12.33) (i) μ T μ ν = 0 , (ii) T , μ μ ν = 0 , (iii) T = 0 , (12.33)  (i)  μ T μ ν = 0 ,  (ii)  T , μ μ ν = 0 ,  (iii)  T = 0 , {:(12.33)" (i) "del_(mu)T^(mu nu)=0","quad" (ii) "T_(,mu)^(mu nu)=0","quad" (iii) "grad*T=0",":}\begin{equation*} \text { (i) } \partial_{\mu} T^{\mu \nu}=0, \quad \text { (ii) } T_{, \mu}^{\mu \nu}=0, \quad \text { (iii) } \boldsymbol{\nabla} \cdot \boldsymbol{T}=0, \tag{12.33} \end{equation*}(12.33) (i) μTμν=0, (ii) T,μμν=0, (iii) T=0,
as justified in the following example.

Example 12.8

  • Putting ν = 0 ν = 0 nu=0\nu=0ν=0 into T μ ν , μ = 0 T μ ν , μ = 0 T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0Tμν,μ=0 gives T 00 , 0 + T 0 i , i = 0 T 00 , 0 + T 0 i , i = 0 T^(00)_(,0)+T^(0i)_(,i)=0T^{00}{ }_{, 0}+T^{0 i}{ }_{, i}=0T00,0+T0i,i=0, which expands into ρ / t + ( ρ v ) = 0 ρ / t + ( ρ v ) = 0 del rho//del t+ vec(grad)*(rho vec(v))=0\partial \rho / \partial t+\vec{\nabla} \cdot(\rho \vec{v})=0ρ/t+(ρv)=0, which is the conservation law given in eqn 12.30 . This part therefore expresses local conservation of energy in flat spacetime.
  • Putting ν = i ν = i nu=i\nu=iν=i into T μ ν , μ = 0 T μ ν , μ = 0 T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0Tμν,μ=0 gives T i 0 , 0 + T i j , j = 0 T i 0 , 0 + T i j , j = 0 T^(i0)_(,0)+T^(ij)_(,j)=0T^{i 0}{ }_{, 0}+T^{i j}{ }_{, j}=0Ti0,0+Tij,j=0, which expands into ( ρ v i ) / t + ( ρ v i v ) = 0 ρ v i / t + ρ v i v = 0 del(rhov^(i))//del t+ vec(grad)*(rhov^(i)( vec(v)))=0\partial\left(\rho v^{i}\right) / \partial t+\vec{\nabla} \cdot\left(\rho v^{i} \vec{v}\right)=0(ρvi)/t+(ρviv)=0, which is a very similar conservation law, but now the i i iii th component of the momentum density ρ v i ρ v i rhov^(i)\rho v^{i}ρvi has replaced the energy. This part therefore expresses local conservation of momentum in flat spacetime.
All this has been for flat spacetime. The next natural step is check if a version of T μ ν , μ = 0 T μ ν , μ = 0 T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0Tμν,μ=0 is valid for curved spacetime. If an observer makes measurements in a local inertial frame (LIF), they will conclude that
(12.34) T , μ ^ μ ^ ν ^ = 0 (12.34) T , μ ^ μ ^ ν ^ = 0 {:(12.34)T_(, hat(mu))^( hat(mu) hat(nu))=0:}\begin{equation*} T_{, \hat{\mu}}^{\hat{\mu} \hat{\nu}}=0 \tag{12.34} \end{equation*}(12.34)T,μ^μ^ν^=0
Recall that, in a LIF, the connection coefficients Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ vanish at a point, 15 15 ^(15){ }^{15}15 making ordinary and covariant derivatives identical. As a result, the observer will conclude that
(12.35) T ; μ ^ μ ^ ν ^ = 0 (12.35) T ; μ ^ μ ^ ν ^ = 0 {:(12.35)T_(; hat(mu))^( hat(mu) hat(nu))=0:}\begin{equation*} T_{; \hat{\mu}}^{\hat{\mu} \hat{\nu}}=0 \tag{12.35} \end{equation*}(12.35)T;μ^μ^ν^=0
This is a valid tensor equation in flat space so, by the principle of general covariance, should also work in curved space, if properly upgraded. The upgrade simply involves removing the hats
(12.36) T ; μ μ ν = 0 (12.36) T ; μ μ ν = 0 {:(12.36)T_(;mu)^(mu nu)=0:}\begin{equation*} T_{; \mu}^{\mu \nu}=0 \tag{12.36} \end{equation*}(12.36)T;μμν=0
and we now have our curved space version. This argument, based on the principle of general covariance, is often called comma goes to semicolon and is equivalent to the swap η g η g eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}ηg. Probably the most memorable way to write this important result, eqn 12.36 , is 16 16 ^(16){ }^{16}16
(12.38) T = 0 (12.38) T = 0 {:(12.38)grad*T=0:}\begin{equation*} \nabla \cdot T=0 \tag{12.38} \end{equation*}(12.38)T=0
Example 12.9
We can take the divergence of the perfect fluid tensor part by part by noting that the covariant derivative obeys the Leibniz ( -=\equiv chain) rule. Start with components T μ ν = ( ρ + p ) u μ u ν + p g μ ν T μ ν = ( ρ + p ) u μ u ν + p g μ ν T^(mu nu)=(rho+p)u^(mu)u^(nu)+pg^(mu nu)T^{\mu \nu}=(\rho+p) u^{\mu} u^{\nu}+p g^{\mu \nu}Tμν=(ρ+p)uμuν+pgμν and using the semicolon notation and the Leibniz rule, we find 17 17 ^(17){ }^{17}17
T ; ν μ ν = ( ρ + p ) ; ν u μ u ν + ( ρ + p ) u ; ν μ u ν + ( ρ + p ) u μ u ; ν ν + p ; ν g μ ν + p g ; ν μ ν . ( 12.39 ) T ; ν μ ν = ( ρ + p ) ; ν u μ u ν + ( ρ + p ) u ; ν μ u ν + ( ρ + p ) u μ u ; ν ν + p ; ν g μ ν + p g ; ν μ ν . ( 12.39 ) T_(;nu)^(mu nu)=(rho+p)_(;nu)u^(mu)u^(nu)+(rho+p)u_(;nu)^(mu)u^(nu)+(rho+p)u^(mu)u_(;nu)^(nu)+p_(;nu)g^(mu nu)+pg_(;nu)^(mu nu).(12.39)T_{; \nu}^{\mu \nu}=(\rho+p)_{; \nu} u^{\mu} u^{\nu}+(\rho+p) u_{; \nu}^{\mu} u^{\nu}+(\rho+p) u^{\mu} u_{; \nu}^{\nu}+p_{; \nu} g^{\mu \nu}+p g_{; \nu}^{\mu \nu} .(12.39)T;νμν=(ρ+p);νuμuν+(ρ+p)u;νμuν+(ρ+p)uμu;νν+p;νgμν+pg;νμν.(12.39)
15 15 ^(15){ }^{15}15 This argument is the same one we rehearsed in Example 8.8.
16 16 ^(16){ }^{16}16 We have not yet worried about how to take the covariant derivative of anything more complicated than a vector. We return to this in Part V, but for now note that, written explicitly, the component equation is
T ; μ μ ν = T μ ν x μ + Γ μ α μ T α ν (12.37) + Γ ν μ α T μ α T ; μ μ ν = T μ ν x μ + Γ μ α μ T α ν (12.37) + Γ ν μ α T μ α {:[T_(;mu)^(mu nu)=(delT^(mu nu))/(delx^(mu))+Gamma_(mu alpha)^(mu)T^(alpha nu)],[(12.37)+Gamma^(nu)_(mu alpha)T^(mu alpha)]:}\begin{align*} T_{; \mu}^{\mu \nu}= & \frac{\partial T^{\mu \nu}}{\partial x^{\mu}}+\Gamma_{\mu \alpha}^{\mu} T^{\alpha \nu} \\ & +\Gamma^{\nu}{ }_{\mu \alpha} T^{\mu \alpha} \tag{12.37} \end{align*}T;μμν=Tμνxμ+ΓμαμTαν(12.37)+ΓνμαTμα
17 17 ^(17){ }^{17}17 Remember that the tensor T T T\boldsymbol{T}T is symmetric, so T μ ν = T ν μ T μ ν = T ν μ T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}Tμν=Tνμ and there is no metric, so T μ ν = T ν μ T μ ν = T ν μ T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}Tμν=Tνμ and there is no
difference if we take the derivative with respect to μ μ mu\muμ or ν ν nu\nuν. Note also that (i) for a scalar function f f fff we have
f ; μ = f , μ f ; μ = f , μ f_(;mu)=f_(,mu)f_{; \mu}=f_{, \mu}f;μ=f,μ
and (ii) we have the important equation
g μ ν ; μ = 0 g μ ν ; μ = 0 g^(mu nu)_(;mu)=0g^{\mu \nu}{ }_{; \mu}=0gμν;μ=0
18 18 ^(18){ }^{18}18 This can be reduced to the geodesi equation in the case of constant pres sure and conserved mass density. See the exercises at the end of this chapter and also Chapter 39.
19 19 ^(19){ }^{19}19 As always, the geometrical interpre tation of velocity is the vector tangent to the world line of a observer.
Fig. 12.4 The world lines of three parallel observers in flat spacetime. With respect to the coordinate frame of the figure, they all have identical velocity.
This quantity must, of course, vanish since T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. Noting the comments in the sidenote, we find
(12.40) 0 = T ; ν μ ν = [ ( ρ + p ) , ν u ν + ( ρ + p ) u ; ν ν ] u μ + ( ρ + p ) u ν u μ ; ν + p , ν g μ ν . (12.40) 0 = T ; ν μ ν = ( ρ + p ) , ν u ν + ( ρ + p ) u ; ν ν u μ + ( ρ + p ) u ν u μ ; ν + p , ν g μ ν . {:(12.40)0=T_(;nu)^(mu nu)=[(rho+p)_(,nu)u^(nu)+(rho+p)u_(;nu)^(nu)]u^(mu)+(rho+p)u^(nu)u^(mu)_(;nu)+p_(,nu)g^(mu nu).:}\begin{equation*} 0=T_{; \nu}^{\mu \nu}=\left[(\rho+p)_{, \nu} u^{\nu}+(\rho+p) u_{; \nu}^{\nu}\right] u^{\mu}+(\rho+p) u^{\nu} u^{\mu}{ }_{; \nu}+p_{, \nu} g^{\mu \nu} . \tag{12.40} \end{equation*}(12.40)0=T;νμν=[(ρ+p),νuν+(ρ+p)u;νν]uμ+(ρ+p)uνuμ;ν+p,νgμν.
We can write this in tensor notation as
(12.41) 0 = T = [ u ρ + u p + ( ρ + p ) ( u ) ] u + ( ρ + p ) u u + p . (12.41) 0 = T = u ρ + u p + ( ρ + p ) ( u ) u + ( ρ + p ) u u + p . {:(12.41)0=grad*T=[grad_(u)rho+grad_(u)p+(rho+p)(grad*u)]u+(rho+p)grad_(u)u+grad p.:}\begin{equation*} 0=\boldsymbol{\nabla} \cdot \boldsymbol{T}=\left[\boldsymbol{\nabla}_{\boldsymbol{u}} \rho+\boldsymbol{\nabla}_{\boldsymbol{u}} p+(\rho+p)(\boldsymbol{\nabla} \cdot \boldsymbol{u})\right] \boldsymbol{u}+(\rho+p) \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}+\boldsymbol{\nabla} p . \tag{12.41} \end{equation*}(12.41)0=T=[uρ+up+(ρ+p)(u)]u+(ρ+p)uu+p.
We shall see in Chapter 39 that this complicated expression gives us an equation of motion for the fluid. 18 18 ^(18){ }^{18}18
We therefore have that T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 is a valid constraint on the energymomentum tensor T T T\boldsymbol{T}T in curved space. It's tempting to interpret T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 as an equation of energy conservation for curved spacetime, but we need to be careful. The conservation of energy over anything other than small distances in curved spacetime is a knotty problem.
To see this, let's first discuss what we mean by the energy E E EEE in general relativity. We define the energy of an object with momentum 1 -form p ~ p ~ tilde(p)\tilde{\boldsymbol{p}}p~ via E = p ~ ( u ) E = p ~ ( u ) E=- tilde(p)(u)E=-\tilde{\boldsymbol{p}}(\boldsymbol{u})E=p~(u), where u u u\boldsymbol{u}u is the velocity 19 19 ^(19){ }^{19}19 of an observer who is located at the site of the object whose energy we are measuring. In flat space, we then define the energy measured by an observer who isn't present, as being equal to the energy measured by the local observer if their velocity is parallel to that of the distant observer. Geometrically, parallel observers have no relative velocity in flat spacetime, so their world lines are parallel straight lines (Fig. 12.4). We can define a velocity field u ( x ) u ( x ) u(x)\boldsymbol{u}(x)u(x) for all observers, which when we input a position in spacetime, will output a vector tangent to the world line of whichever observer is at that spacetime point. Mathematically, the condition that these flat-spacetime observers have parallel velocity may be written as
(12.42) u , μ ν = 0 (12.42) u , μ ν = 0 {:(12.42)u_(,mu)^(nu)=0:}\begin{equation*} u_{, \mu}^{\nu}=0 \tag{12.42} \end{equation*}(12.42)u,μν=0
which tells us that the tangent vectors to the world lines don't vary in space, and so are parallel. The mathematical need for parallel observers defined in this way is justified in the next example.
Example 12.10
In flat spacetime, the vanishing divergence is written as T μ ν , μ = 0 T μ ν , μ = 0 T^(mu nu),mu=0T^{\mu \nu}, \mu=0Tμν,μ=0. Recall that for dust we had T = J ~ p ~ T = J ~ p ~ T= tilde(J)ox tilde(p)\boldsymbol{T}=\tilde{\boldsymbol{J}} \otimes \tilde{\boldsymbol{p}}T=J~p~ and so the energy density can be written as
(12.43) T ( u , u ) = T μ ν u μ u ν = E J ~ ( u ) (12.43) T ( u , u ) = T μ ν u μ u ν = E J ~ ( u ) {:(12.43)-T(u","u)=-T_(mu nu)u^(mu)u^(nu)=E tilde(J)(u):}\begin{equation*} -\boldsymbol{T}(\boldsymbol{u}, \boldsymbol{u})=-T_{\mu \nu} u^{\mu} u^{\nu}=E \tilde{\boldsymbol{J}}(\boldsymbol{u}) \tag{12.43} \end{equation*}(12.43)T(u,u)=Tμνuμuν=EJ~(u)
The quantity M ~ ( ) = E J ~ ( ) M ~ ( ) = E J ~ ( ) tilde(M)()=E tilde(J)()\tilde{\boldsymbol{M}}()=E \tilde{\boldsymbol{J}}()M~()=EJ~() can be thought of as the current 1-form of energy carried by the particles in the dust cloud. If this is to be locally conserved, we must hav
M μ , μ = 0 M μ , μ = 0 M^(mu)_(,mu)=0M^{\mu}{ }_{, \mu}=0Mμ,μ=0, where M μ M μ M^(mu)M^{\mu}Mμ are the components of vector M M M\boldsymbol{M}M which, from eqn 12.43 are
(12.44) M μ = T μ ν u ν (12.44) M μ = T μ ν u ν {:(12.44)M^(mu)=-T^(mu nu)u_(nu):}\begin{equation*} M^{\mu}=-T^{\mu \nu} u_{\nu} \tag{12.44} \end{equation*}(12.44)Mμ=Tμνuν
The condition that M μ , μ = 0 M μ , μ = 0 M^(mu)_(,mu)=0M^{\mu}{ }_{, \mu}=0Mμ,μ=0 therefore requires that the velocity field obeys
(12.45) 0 = ( T μ ν u ν ) , μ = ( T , μ μ ν ) u ν T μ ν ( u ν , μ ) = T μ ν u ν , μ (12.45) 0 = T μ ν u ν , μ = T , μ μ ν u ν T μ ν u ν , μ = T μ ν u ν , μ {:(12.45)0=(-T^(mu nu)u_(nu))_(,mu)=-(T_(,mu)^(mu nu))u_(nu)-T^(mu nu)(u_(nu,mu))=-T^(mu nu)u_(nu,mu):}\begin{equation*} 0=\left(-T^{\mu \nu} u_{\nu}\right)_{, \mu}=-\left(T_{, \mu}^{\mu \nu}\right) u_{\nu}-T^{\mu \nu}\left(u_{\nu, \mu}\right)=-T^{\mu \nu} u_{\nu, \mu} \tag{12.45} \end{equation*}(12.45)0=(Tμνuν),μ=(T,μμν)uνTμν(uν,μ)=Tμνuν,μ
where we have used the Leibniz rule and T μ ν , μ = 0 T μ ν , μ = 0 T^(mu nu)_(,mu)=0T^{\mu \nu}{ }_{, \mu}=0Tμν,μ=0. To guarantee that this vanishes, we require u ν , μ = 0 u ν , μ = 0 u^(nu)_(,mu)=0u^{\nu}{ }_{, \mu}=0uν,μ=0. Physically, this requires us to have access to a family of inertial, parallel observers if we are to have a notion of conservation of energy.
The argument above is appropriate for flat spacetime. However, in curved spacetime we lose the ability to globally define a family of parallel observers, and this will frustrate our attempts to interpret T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 as statement of energy conservation. This is because 'being parallel' is a path-dependent property in curved spacetime, as we see in the next example. 20 20 ^(20){ }^{20}20
Example 12.11
Replaying the last example for curved spacetime, we apply a condition T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 using the covariant derivative, and so we need
(12.46) ( T μ ν u ν ) ; μ = 0 (12.46) T μ ν u ν ; μ = 0 {:(12.46)(T^(mu nu)u_(nu))_(;mu)=0:}\begin{equation*} \left(T^{\mu \nu} u_{\nu}\right)_{; \mu}=0 \tag{12.46} \end{equation*}(12.46)(Tμνuν);μ=0
Follow the same steps as before, we conclude that this is so if u ν ; μ = 0 u ν ; μ = 0 u_(nu;mu)=0u_{\nu ; \mu}=0uν;μ=0, which, owing to the symmetry of T T T\boldsymbol{T}T, can be rewritten as
(12.47) u ν ; μ + u μ ; ν = 0 . (12.47) u ν ; μ + u μ ; ν = 0 . {:(12.47)u_(nu;mu)+u_(mu;nu)=0.:}\begin{equation*} u_{\nu ; \mu}+u_{\mu ; \nu}=0 . \tag{12.47} \end{equation*}(12.47)uν;μ+uμ;ν=0.
This condition is known as Killing's equation (Chapter 33) and is not generally true of velocity vector fields in curved spacetime. 21 21 ^(21){ }^{21}21
The previous example implies that in curved spacetime, although T = T = grad*T=\boldsymbol{\nabla} \cdot \boldsymbol{T}=T= 0 holds, it can't be strictly interpreted in terms of energy conservation. Physically, this is to be expected since general relativity shows that matter and gravity are coupled: the gravitational field (described by the curvature of spacetime) can do work on matter, and matter can do work on the gravitational field. However, over small distances they don't do much work on each other and so, approximately at least, we have energy conservation 22 22 ^(22){ }^{22}22 since u ν ; μ 0 u ν ; μ 0 u_(nu;mu)~~0u_{\nu ; \mu} \approx 0uν;μ0.
We now have the object T T T\boldsymbol{T}T that, in some form, lives on the right-hand side of Einstein's equation, along with its key property 23 T = 0 23 T = 0 ^(23)grad*T=0{ }^{23} \boldsymbol{\nabla} \cdot \boldsymbol{T}=023T=0 (which is true, in spite of its rather subtle interpretation). Our next task is to marry the geometric left-hand side of Einstein's equation with the physical right-hand side.

Chapter summary

  • The right-hand side of Einstein's equation encodes the energymomentum of matter fields using the tensor T T T\boldsymbol{T}T.
  • In flat spacetime, the local conservation of energy-momentum can be expressed via the important constraint T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. The same expression holds in curved spacetime.
    20 20 ^(20){ }^{20}20 Another way of seeing this point is that, to globally conserve momentum, we need to demonstrate
V T μ ν d Σ ν = 0 V T μ ν d Σ ν = 0 int_(delV)T^(mu nu)dSigma_(nu)=0\int_{\partial \mathcal{V}} T^{\mu \nu} \mathrm{d} \Sigma_{\nu}=0VTμνdΣν=0
where V V delV\partial \mathcal{V}V is the boundary of some closed region of spacetime V V V\mathcal{V}V. But to work this out you have to combine lots of individual 4 -vectors T μ ν d Σ ν T μ ν d Σ ν T^(mu nu)dSigma_(nu)T^{\mu \nu} \mathrm{d} \Sigma_{\nu}TμνdΣν from different parts of the surface and bring them to a common location to add them to a com add them all up. No problem in flat space-
time, but in curved spacetime they all live in different tangent spaces and this live in different tangent spaces and this
process cannot be done in a well-defined manner.
21 21 ^(21){ }^{21}21 Special cases that do obey Killing's equation are very interesting and we exequation are very interesting and we ex-
amine them in detail from Part IV onwards.
22 22 ^(22){ }^{22}22 We shall therefore call T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 a conservation equation in this book, in a slight abuse of language. The law T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 can be proven geometrically from the invariance of the manifold with respect to a very general type of translation known as a diffeomorphism. (See Appendix C for more details.) So although it does not represent conservation of energy in the strict sense, T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 is still a significant constraint on our equations.
23 23 ^(23){ }^{23}23 We also remark that T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 is connected with the geodesic principle, namely that free massive point particles follow timelike geodesics. The conservation equation captures the idea that the massive particle is free, that is, it is not exchanging energy-momentum with the fields and matter in its environment. Together with an energy condition (see page 145 for more details) to capture the idea that energy propagates within the body in a timelike or null manner, the geodesic principle can be proved (see e.g. J. Ehlers and R. Geroch, Ann. Phys. 309, 232 (2004) for a proof, and J. O. Weatherall, Studies in the History and Philosophy of Modern Physics 42, 276 (2011) for a reflection Physics 42, 276 (2011) for a reflection
on the debate that has ensued about on the debate that has ens
the status of the principle).

Exercises

(12.1) Consider the transformations between Cartesian coordinates ( t , x , y , z ) ( t , x , y , z ) (t,x,y,z)(t, x, y, z)(t,x,y,z) and spherical polar coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ). In the Cartesian system, the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor T T T\boldsymbol{T}T is diagonal with non-zero components
(12.48) T t t = ρ , T i i = p (12.48) T t t = ρ , T i i = p {:(12.48)T_(tt)=rho","quadT_(ii)=p:}\begin{equation*} T_{t t}=\rho, \quad T_{i i}=p \tag{12.48} \end{equation*}(12.48)Ttt=ρ,Tii=p
where i = x , y , z i = x , y , z i=x,y,zi=x, y, zi=x,y,z. Transform the components of this tensor to spherical polars.
(12.2) Consider dust with energy density T T T\boldsymbol{T}T with components T μ ν = ρ 0 u μ u ν T μ ν = ρ 0 u μ u ν T^(mu nu)=rho_(0)u^(mu)u^(nu)T^{\mu \nu}=\rho_{0} u^{\mu} u^{\nu}Tμν=ρ0uμuν.
(a) Explain why ( ρ 0 u ) = 0 ρ 0 u = 0 grad*(rho_(0)u)=0\boldsymbol{\nabla} \cdot\left(\rho_{0} \boldsymbol{u}\right)=0(ρ0u)=0.
(b) Show that
(12.49) 0 = T = u μ ( ρ 0 u ν ) ; ν + ρ 0 u ν u ; ν μ (12.49) 0 = T = u μ ρ 0 u ν ; ν + ρ 0 u ν u ; ν μ {:(12.49)0=grad*T=u^(mu)(rho_(0)u^(nu))_(;nu)+rho_(0)u^(nu)u_(;nu)^(mu):}\begin{equation*} 0=\boldsymbol{\nabla} \cdot \boldsymbol{T}=u^{\mu}\left(\rho_{0} u^{\nu}\right)_{; \nu}+\rho_{0} u^{\nu} u_{; \nu}^{\mu} \tag{12.49} \end{equation*}(12.49)0=T=uμ(ρ0uν);ν+ρ0uνu;νμ
(c) Using the result of part (a) show that the con servation of mass-energy guarantees that the dust particles obey the geodesic equation and hence that dust particles follow geodesic world lines.
(12.3) We write an energy-momentum tensor for a particle as
(12.50) T μ ν ( x ) = d τ m g d z μ d τ d z ν d τ δ ( 4 ) [ x z ( τ ) ] (12.50) T μ ν ( x ) = d τ m g d z μ d τ d z ν d τ δ ( 4 ) [ x z ( τ ) ] {:(12.50)T^(mu nu)(x)=intdtau(m)/(sqrt(-g))((d)z^(mu))/(dtau)((d)z^(nu))/(dtau)*delta^((4))[x-z(tau)]:}\begin{equation*} T^{\mu \nu}(x)=\int \mathrm{d} \tau \frac{m}{\sqrt{-g}} \frac{\mathrm{~d} z^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{~d} z^{\nu}}{\mathrm{d} \tau} \cdot \delta^{(4)}[x-z(\tau)] \tag{12.50} \end{equation*}(12.50)Tμν(x)=dτmg dzμdτ dzνdτδ(4)[xz(τ)]
where z ( τ ) z ( τ ) z(tau)z(\tau)z(τ) is the world line of the particle parametrized by proper time τ τ tau\tauτ. A useful expression when using this tensor, is that for functions f ( x ) f ( x ) f(x)f(x)f(x) and g ( x ) g ( x ) g(x)g(x)g(x) we have
(12.51) d x δ [ f ( x ) ] g ( x ) = a g ( x a ) | f ( x a ) | (12.51) d x δ [ f ( x ) ] g ( x ) = a g x a f x a {:(12.51)intdx delta[f(x)]g(x)=sum_(a)(g(x_(a)))/(|f^(')(x_(a))|):}\begin{equation*} \int \mathrm{d} x \delta[f(x)] g(x)=\sum_{a} \frac{g\left(x_{a}\right)}{\left|f^{\prime}\left(x_{a}\right)\right|} \tag{12.51} \end{equation*}(12.51)dxδ[f(x)]g(x)=ag(xa)|f(xa)|
where the sum over a a aaa is over all values of x a x a x_(a)x_{a}xa that have the property that f ( x a ) = 0 f x a = 0 f(x_(a))=0f\left(x_{a}\right)=0f(xa)=0.
Using the tools above, justify the following expressions for the components of the energy-momentum tensor for a swarm of particles in flat space.
(a) The momentum density
(12.52) T α 0 ( x ) = n p n α ( τ n ) δ ( 3 ) [ x z n ( τ n ) ] (12.52) T α 0 ( x ) = n p n α τ n δ ( 3 ) x z n τ n {:(12.52)T^(alpha0)(x)=sum_(n)p_(n)^(alpha)(tau_(n))delta^((3))[x-z_(n)(tau_(n))]:}\begin{equation*} T^{\alpha 0}(x)=\sum_{n} p_{n}^{\alpha}\left(\tau_{n}\right) \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \tag{12.52} \end{equation*}(12.52)Tα0(x)=npnα(τn)δ(3)[xzn(τn)]
(b) The momentum current
(12.53) T α i ( x ) = n p n α ( τ n ) d z n i ( τ n ) d t δ ( 3 ) [ x z n ( τ n ) ] ; (12.53) T α i ( x ) = n p n α τ n d z n i τ n d t δ ( 3 ) x z n τ n ; {:(12.53)T^(alpha i)(x)=sum_(n)p_(n)^(alpha)(tau_(n))(dz_(n)^(i)(tau_(n)))/(dt)delta^((3))[x-z_(n)(tau_(n))];:}\begin{equation*} T^{\alpha i}(x)=\sum_{n} p_{n}^{\alpha}\left(\tau_{n}\right) \frac{\mathrm{d} z_{n}^{i}\left(\tau_{n}\right)}{\mathrm{d} t} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] ; \tag{12.53} \end{equation*}(12.53)Tαi(x)=npnα(τn)dzni(τn)dtδ(3)[xzn(τn)];
(c) The energy-momentum tensor
(12.54) T α β ( x ) = n p n α ( τ n ) p n β ( τ n ) E n ( τ n ) δ ( 3 ) [ x z n ( τ n ) ] (12.54) T α β ( x ) = n p n α τ n p n β τ n E n τ n δ ( 3 ) x z n τ n {:(12.54)T^(alpha beta)(x)=sum_(n)(p_(n)^(alpha)(tau_(n))p_(n)^(beta)(tau_(n)))/(E_(n)(tau_(n)))delta^((3))[x-z_(n)(tau_(n))]:}\begin{equation*} T^{\alpha \beta}(x)=\sum_{n} \frac{p_{n}^{\alpha}\left(\tau_{n}\right) p_{n}^{\beta}\left(\tau_{n}\right)}{E_{n}\left(\tau_{n}\right)} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \tag{12.54} \end{equation*}(12.54)Tαβ(x)=npnα(τn)pnβ(τn)En(τn)δ(3)[xzn(τn)]
where the index n n nnn labels a specific particle and τ n τ n tau_(n)\tau_{n}τn solves the equation z 0 ( τ n ) = t z 0 τ n = t z^(0)(tau_(n))=tz^{0}\left(\tau_{n}\right)=tz0(τn)=t and t t ttt is the coordinate time [i.e. particle n n nnn 's proper time at the coordinate time t t ttt that is inputted into the tensor T ( x ) T ( x ) T(x)T(x)T(x) as part of the argument x x xxx with components x μ = ( t , x ) ] x μ = ( t , x ) {:x^(mu)=(t,( vec(x)))]\left.x^{\mu}=(t, \vec{x})\right]xμ=(t,x)].
(12.4) By applying T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 to the energy-momentum tensor for a single particle in the previous question, show that we obtain the geodesic equation.
(12.5) In flat space, define a ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) energy-momentum tensor for dust particles as
(12.55) T ( , ) = p J ~ (12.55) T ( , ) = p J ~ {:(12.55)T(",")=p ox tilde(J):}\begin{equation*} \boldsymbol{T}(,)=\boldsymbol{p} \otimes \tilde{\boldsymbol{J}} \tag{12.55} \end{equation*}(12.55)T(,)=pJ~
This ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) object has two slots: one for a 1 -form and one for a vector. It has components T μ ν T μ ν T^(mu)_(nu)T^{\mu}{ }_{\nu}Tμν.
(a) Find the eigenvalue of this tensor when the velocity eigenvector v v v\boldsymbol{v}v is inserted. That is, determine the constants α α alpha\alphaα for the equation
(12.56) T ( , v ) = α v (12.56) T ( , v ) = α v {:(12.56)T(","v)=alpha v:}\begin{equation*} \boldsymbol{T}(, \boldsymbol{v})=\alpha \boldsymbol{v} \tag{12.56} \end{equation*}(12.56)T(,v)=αv
Note that the other three eigenvectors are orthogonal to v v v\boldsymbol{v}v and therefore in the 3-space we can call v v v^(_|_)\boldsymbol{v}^{\perp}v By isotropy these three eigenvectors are degenerate and have some eigenvalue p p ppp, such that
(12.57) T ( , e i ) = p e i . (12.57) T , e i = p e i . {:(12.57)T(,e_(i))=pe_(i).:}\begin{equation*} \boldsymbol{T}\left(, \boldsymbol{e}_{i}\right)=p \boldsymbol{e}_{i} . \tag{12.57} \end{equation*}(12.57)T(,ei)=pei.
(b) Show that we can write
(12.58) T ( Y ~ , X ) = X 0 ^ Y 0 ^ ( ρ + p ) + X μ ^ Y μ ^ p . (12.58) T ( Y ~ , X ) = X 0 ^ Y 0 ^ ( ρ + p ) + X μ ^ Y μ ^ p . {:(12.58)T( tilde(Y)","X)=-X^( hat(0))Y_( hat(0))(rho+p)+X^( hat(mu))Y_( hat(mu))p.:}\begin{equation*} \boldsymbol{T}(\tilde{\boldsymbol{Y}}, \boldsymbol{X})=-X^{\hat{0}} Y_{\hat{0}}(\rho+p)+X^{\hat{\mu}} Y_{\hat{\mu}} p . \tag{12.58} \end{equation*}(12.58)T(Y~,X)=X0^Y0^(ρ+p)+Xμ^Yμ^p.
(c) Use the result of part (b) result to justify the curved space expression
(12.59) T μ ν = ( ρ + p ) v μ v ν + p g μ ν (12.59) T μ ν = ( ρ + p ) v μ v ν + p g μ ν {:(12.59)T_(mu nu)=(rho+p)v_(mu)v_(nu)+pg_(mu nu):}\begin{equation*} T_{\mu \nu}=(\rho+p) v_{\mu} v_{\nu}+p g_{\mu \nu} \tag{12.59} \end{equation*}(12.59)Tμν=(ρ+p)vμvν+pgμν
See the book by Ludvigsen for a discussion of this approach.

The gravitational field equations

The key idea of general relativity is that the curvature of spacetime, which gives rise to gravitation, is determined by the energy density of the matter fields of the Universe, such that we can write the Einstein equation
(13.1) ( Curvature of spacetime ) = ( Energy density of matter fields ) . (13.1) (  Curvature of   spacetime  ) = (  Energy density   of matter fields  ) . {:(13.1)((" Curvature of ")/(" spacetime "))=((" Energy density ")/(" of matter fields ")).:}\begin{equation*} \binom{\text { Curvature of }}{\text { spacetime }}=\binom{\text { Energy density }}{\text { of matter fields }} . \tag{13.1} \end{equation*}(13.1)( Curvature of  spacetime )=( Energy density  of matter fields ).
At last, in this chapter, we are able to formulate the details of this equation! Our task is, on the (geometrical) left-hand side, to identify a suitable expression for the curvature of spacetime and, on the (physical) right, a suitable expression for the density of energy in matter fields.
In order to get to Einstein's equation, we take the route shown in Fig. 13.1. Starting with the physical principles of fields and of eqn 13.1, we shall review the key elements of geometry that supply the left-hand side of the equation. We then examine the physics of the right-hand side, in particular the energy-momentum tensor field T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x) that encodes the energy density of the matter fields that fill the Universe. Finally we link these to form the Einstein field equation, the tensor field version of eqn 13.1.

13.1 Geometry: a recap of the key ingredients

The metric tensor field g ( x ) g ( x ) g(x)\boldsymbol{g}(x)g(x) is the foundation of the curvature of spacetime. As we've written it, it is a classical field: a function of position in spacetime x x xxx that gives us a tensor valid at the point x x xxx. We often express the components of the metric field g μ ν ( x ) g μ ν ( x ) g_(mu nu)(x)g_{\mu \nu}(x)gμν(x) via a line element
(13.2) d s 2 = g μ ν d x μ d x ν (13.2) d s 2 = g μ ν d x μ d x ν {:(13.2)ds^(2)=g_(mu nu)dx^(mu)dx^(nu):}\begin{equation*} \mathrm{d} s^{2}=g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \tag{13.2} \end{equation*}(13.2)ds2=gμνdxμdxν
There is no way to tell whether a single particle accelerates due to the effects of a homogeneous gravitational field or because of the choice of coordinates. If a gravitational field is present, and all real-life gravitational fields are necessarily inhomogeneous, spacetime is curved and this gives rise to the acceleration of a particle in the absence of forces external to spacetime. However, a particle might accelerate merely due to the perversity of the coordinate system that we've chosen, rather than due to curvature. The method to unambiguously detect curvature
13.1 Geometry: a recap of the key ingredients
Fig. 13.1 A conceptual route to Einstein's equation. The left-hand side leads to the Einstein tensor field G G G\boldsymbol{G}G, related to the Riemann tensor field as described later in this chapter. The right-hand side leads to a constant 8 π G 8 π G 8pi G8 \pi G8πG (where G G GGG here is the gravitational constant) multiplied by the energymomentum tensor field T T T\boldsymbol{T}T.
1 1 ^(1){ }^{1}1 The connection coefficients may be re lated to the metric tensor field via the component equation
Γ α β μ = (13.3) 1 2 g μ λ ( g λ α x β + g λ β x α g α β x λ ) Γ α β μ = (13.3) 1 2 g μ λ g λ α x β + g λ β x α g α β x λ {:[Gamma_(alpha beta)^(mu)=],[(13.3)(1)/(2)g^(mu lambda)((delg_(lambda alpha))/(delx^(beta))+(delg_(lambda beta))/(delx^(alpha))-(delg_(alpha beta))/(delx^(lambda)))]:}\begin{align*} & \Gamma_{\alpha \beta}^{\mu}= \\ & \frac{1}{2} g^{\mu \lambda}\left(\frac{\partial g_{\lambda \alpha}}{\partial x^{\beta}}+\frac{\partial g_{\lambda \beta}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}\right) \tag{13.3} \end{align*}Γαβμ=(13.3)12gμλ(gλαxβ+gλβxαgαβxλ)
2 2 ^(2){ }^{2}2 We use
(13.4) D 2 ξ d τ 2 = R ( , u , ξ , u ) , (13.4) D 2 ξ d τ 2 = R ( , u , ξ , u ) , {:(13.4)(D^(2)xi)/((d)tau^(2))=R(","u","xi","u)",":}\begin{equation*} \frac{D^{2} \boldsymbol{\xi}}{\mathrm{~d} \tau^{2}}=\boldsymbol{R}(, \boldsymbol{u}, \boldsymbol{\xi}, \boldsymbol{u}), \tag{13.4} \end{equation*}(13.4)D2ξ dτ2=R(,u,ξ,u),
where ξ ξ xi\boldsymbol{\xi}ξ is the particles' separation and u u u\boldsymbol{u}u is the velocity of the particle follow ing the fiducial geodesic.
3 3 ^(3){ }^{3}3 Note for later in the chapter that this is symmetric: R μ ν ( x ) = R ν μ ( x ) R μ ν ( x ) = R ν μ ( x ) R_(mu nu)(x)=R_(nu mu)(x)R_{\mu \nu}(x)=R_{\nu \mu}(x)Rμν(x)=Rνμ(x).
4 4 ^(4){ }^{4}4 Recall that a scalar is the same, independent of the coordinate system in which it is evaluated.
5 5 ^(5){ }^{5}5 One point of view is that the ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) Rie mann curvature tensor field R ( x ) R ( x ) R(x)\boldsymbol{R}(x)R(x) is the gravitational field. So although we have the equivalence principle telling us that a measurement at a point cannot tell us whether it's gravitation or accelera tion causing an effect, we can identify the physical field that causes gravitation. We cannot simply slot this 'gravitational field R R R\boldsymbol{R}R into the Einstein equation (curvature) = = === (energy), owing to the ( 1,3 ) valence of R R R\boldsymbol{R}R compare to the ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) valence of the energy-momentum (2,0) T T TTT However, if we momentum tensor T T TTT. Hower, if we are looking for the physical manifestation of grav itation, this is as good a candidate a any.
6 6 ^(6){ }^{6}6 See also Section 0.4.
involves transforming away the effects due to the coordinate system to arrive at what's left behind: the curvature due to gravitation. This is the motivation behind the Riemann curvature tensor R ( x ) R ( x ) R(x)\boldsymbol{R}(x)R(x), which evaluates the presence of this stuff that's left behind. We expect that this tensor should play a key role in the left-hand side of the equation, which indeed it does. It depends on the (first and second) derivatives of the metric field. The first derivatives of g ( x ) g ( x ) g(x)\boldsymbol{g}(x)g(x) provide the connection coefficients Γ μ α β ( x ) , 1 Γ μ α β ( x ) , 1 Gamma^(mu)_(alpha beta)(x),^(1)\Gamma^{\mu}{ }_{\alpha \beta}(x),{ }^{1}Γμαβ(x),1 and reflect how the metric causes the coordinates to change as we move through spacetime. They are easily manipulated, so we often describe R ( x ) R ( x ) R(x)\boldsymbol{R}(x)R(x) in terms of them, rather directly through g ( x ) g ( x ) g(x)\boldsymbol{g}(x)g(x).
To obtain R ( x ) R ( x ) R(x)\boldsymbol{R}(x)R(x) and thus determine the curvature of spacetime, we evaluate the geodesic deviation of two freely falling particles whose motion was initially parallel and extract R ( x ) R ( x ) R(x)\boldsymbol{R}(x)R(x) from the equation of motion of their separation. 2 2 ^(2){ }^{2}2 The resulting Riemann tensor field has components
R α β γ δ ( x ) = Γ α δ β ( x ) x γ Γ α γ β ( x ) x δ + Γ α γ μ ( x ) Γ μ δ β ( x ) Γ α δ μ ( x ) Γ μ γ β ( x ) R α β γ δ ( x ) = Γ α δ β ( x ) x γ Γ α γ β ( x ) x δ + Γ α γ μ ( x ) Γ μ δ β ( x ) Γ α δ μ ( x ) Γ μ γ β ( x ) R^(alpha)_(beta gamma delta)(x)=(delGamma^(alpha)_(delta beta)(x))/(delx^(gamma))-(delGamma^(alpha)_(gamma beta)(x))/(delx^(delta))+Gamma^(alpha)_(gamma mu)(x)Gamma^(mu)_(delta beta)(x)-Gamma^(alpha)_(delta mu)(x)Gamma^(mu)_(gamma beta)(x)R^{\alpha}{ }_{\beta \gamma \delta}(x)=\frac{\partial \Gamma^{\alpha}{ }_{\delta \beta}(x)}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\gamma \beta}(x)}{\partial x^{\delta}}+\Gamma^{\alpha}{ }_{\gamma \mu}(x) \Gamma^{\mu}{ }_{\delta \beta}(x)-\Gamma^{\alpha}{ }_{\delta \mu}(x) \Gamma^{\mu}{ }_{\gamma \beta}(x)Rαβγδ(x)=Γαδβ(x)xγΓαγβ(x)xδ+Γαγμ(x)Γμδβ(x)Γαδμ(x)Γμγβ(x).
This gives us access to the curvature at each point in spacetime. Two quantities derived from the Riemann tensor field will be important in setting up the Einstein equation. The first is the Ricci tensor field, which at a spacetime point x x xxx has components 3 3 ^(3){ }^{3}3
(13.6) R μ ν ( x ) = R μ α ν α ( x ) (13.6) R μ ν ( x ) = R μ α ν α ( x ) {:(13.6)R_(mu nu)(x)=R_(mu alpha nu)^(alpha)(x):}\begin{equation*} R_{\mu \nu}(x)=R_{\mu \alpha \nu}^{\alpha}(x) \tag{13.6} \end{equation*}(13.6)Rμν(x)=Rμανα(x)
The second is the Ricci scalar field 4 4 ^(4){ }^{4}4
(13.7) R ( x ) = g μ ν R μ ν ( x ) . (13.7) R ( x ) = g μ ν R μ ν ( x ) . {:(13.7)R(x)=g_(mu nu)R^(mu nu)(x).:}\begin{equation*} R(x)=g_{\mu \nu} R^{\mu \nu}(x) . \tag{13.7} \end{equation*}(13.7)R(x)=gμνRμν(x).
Using these two questions, we will be able to gain access to the curvature of spacetime. This will be achieved by taking a double derivative of the metric field g ( x ) g ( x ) g(x)\boldsymbol{g}(x)g(x), which is the basis of the left-hand side of the Einstein equation. 5 5 ^(5){ }^{5}5 After having reviewed the geometrical left-hand side of the equation, let's now turn to the right-hand side and the physical content of the theory.

13.2 Physics: the key ingredients

We start with a principle: a physical relativistic/geometrical field theory of gravitation must be compatible with Newton's theory in the nonrelativistic limit. It is therefore useful to revisit Newton's law of gravitation from the point of view of fields and see what it teaches us about the shape that our new theory must take. 6 6 ^(6){ }^{6}6 Newton's law says that the force F F vec(F)\vec{F}F experienced by a mass m m mmm from a gravitating object with mass M M MMM separated by a displacement vector r r vec(r)\vec{r}r, is given by
(13.8) F ( r ) = G M m r 3 r , (13.8) F ( r ) = G M m r 3 r , {:(13.8) vec(F)(r)=-(GMm)/(r^(3))* vec(r)",":}\begin{equation*} \vec{F}(r)=-\frac{G M m}{r^{3}} \cdot \vec{r}, \tag{13.8} \end{equation*}(13.8)F(r)=GMmr3r,
where G = 6.672 ( 4 ) × 10 11 m 3 kg 1 s 2 G = 6.672 ( 4 ) × 10 11 m 3 kg 1 s 2 G=6.672(4)xx10^(-11)m^(3)kg^(-1)s^(-2)G=6.672(4) \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}G=6.672(4)×1011 m3 kg1 s2 is the gravitational constant. We also find it useful to deal with a scalar potential energy function
U ( r ) U ( r ) U(r)U(r)U(r) experienced by mass m m mmm due to mass M M MMM and given by
(13.9) U ( r ) = G M m r (13.9) U ( r ) = G M m r {:(13.9)U(r)=-(GMm)/(r):}\begin{equation*} U(r)=-\frac{G M m}{r} \tag{13.9} \end{equation*}(13.9)U(r)=GMmr
where F = U F = U vec(F)=- vec(grad)U\vec{F}=-\vec{\nabla} UF=U. The function is U ( r ) U ( r ) U(r)U(r)U(r) is proportional to m m mmm, so it is better to think about the potential Φ ( r ) = U ( r ) / m Φ ( r ) = U ( r ) / m Phi(r)=U(r)//m\Phi(r)=U(r) / mΦ(r)=U(r)/m, the potential energy per unit mass. Newton's law of gravitation can therefore be turned into a field equation straightforwardly. The field version of the law describes the force per unit mass g ( x ) g ( x ) vec(g)( vec(x))\vec{g}(\vec{x})g(x) experienced at a position x x vec(x)\vec{x}x due to the presence of a distribution of mass with density ρ ( x ) ρ x rho( vec(x)^('))\rho\left(\vec{x}^{\prime}\right)ρ(x). The generalization of the force law is
(13.10) g ( x ) = G d 3 x ρ ( x ) ( x x ) | x x | 3 (13.10) g ( x ) = G d 3 x ρ x x x x x 3 {:(13.10) vec(g)( vec(x))=-G intd^(3)x^(')rho( vec(x)^('))((( vec(x))- vec(x)^(')))/(|( vec(x))- vec(x)^(')|^(3)):}\begin{equation*} \vec{g}(\vec{x})=-G \int \mathrm{~d}^{3} x^{\prime} \rho\left(\vec{x}^{\prime}\right) \frac{\left(\vec{x}-\vec{x}^{\prime}\right)}{\left|\vec{x}-\vec{x}^{\prime}\right|^{3}} \tag{13.10} \end{equation*}(13.10)g(x)=G d3xρ(x)(xx)|xx|3
We also define a gravitational potential
(13.11) Φ ( x ) = G d 3 x ρ ( x ) | x x | (13.11) Φ ( x ) = G d 3 x ρ x x x {:(13.11)Phi( vec(x))=-G intd^(3)x^(')(rho( vec(x)^(')))/(|( vec(x))- vec(x)^(')|):}\begin{equation*} \Phi(\vec{x})=-G \int \mathrm{~d}^{3} x^{\prime} \frac{\rho\left(\vec{x}^{\prime}\right)}{\left|\vec{x}-\vec{x}^{\prime}\right|} \tag{13.11} \end{equation*}(13.11)Φ(x)=G d3xρ(x)|xx|
We can then show that (i) the function Φ Φ Phi\PhiΦ obeys the important rule g ( x ) = Φ ( x ) g ( x ) = Φ ( x ) vec(g)( vec(x))=-grad Phi( vec(x))\vec{g}(\vec{x})=-\nabla \Phi(\vec{x})g(x)=Φ(x), and, most importantly, (ii) Φ Φ Phi\PhiΦ can be computed from a given distribution of mass. This is the purpose of the next example.

Example 13.1

(i) Using the potential Φ ( x ) Φ ( x ) Phi( vec(x))\Phi(\vec{x})Φ(x) we can write 7 7 ^(7){ }^{7}7
(13.13) g ( x ) = x d 3 x G ρ ( x ) | x x | = Φ ( x ) (13.13) g ( x ) = x d 3 x G ρ x x x = Φ ( x ) {:(13.13) vec(g)( vec(x))= vec(grad)_( vec(x))intd^(3)x^(')(G rho( vec(x)^(')))/(|( vec(x))- vec(x)^(')|)=- vec(grad)Phi( vec(x)):}\begin{equation*} \vec{g}(\vec{x})=\vec{\nabla}_{\vec{x}} \int \mathrm{~d}^{3} x^{\prime} \frac{G \rho\left(\vec{x}^{\prime}\right)}{\left|\vec{x}-\vec{x}^{\prime}\right|}=-\vec{\nabla} \Phi(\vec{x}) \tag{13.13} \end{equation*}(13.13)g(x)=x d3xGρ(x)|xx|=Φ(x)
(ii) Next, we use the result 8 8 ^(8){ }^{8}8
(13.15) x ( x x | x x | 3 ) = 4 π δ ( 3 ) ( x x ) (13.15) x x x x x 3 = 4 π δ ( 3 ) x x {:(13.15) vec(grad)_( vec(x))*((( vec(x))- vec(x)^('))/(|( vec(x))- vec(x)^(')|^(3)))=4pidelta^((3))(( vec(x))- vec(x)^(')):}\begin{equation*} \vec{\nabla}_{\vec{x}} \cdot\left(\frac{\vec{x}-\vec{x}^{\prime}}{\left|\vec{x}-\vec{x}^{\prime}\right|^{3}}\right)=4 \pi \delta^{(3)}\left(\vec{x}-\vec{x}^{\prime}\right) \tag{13.15} \end{equation*}(13.15)x(xx|xx|3)=4πδ(3)(xx)
which allows us to take the divergence of g g vec(g)\vec{g}g in eqn 13.10 to find
g ( x ) = 4 π G d 3 x ρ ( x ) δ ( 3 ) ( x x ) (13.16) = 4 π G ρ ( x ) g ( x ) = 4 π G d 3 x ρ x δ ( 3 ) x x (13.16) = 4 π G ρ ( x ) {:[ vec(grad)*g( vec(x))=-4pi G intd^(3)x^(')rho( vec(x)^('))delta^((3))(( vec(x))- vec(x)^('))],[(13.16)=-4pi G rho( vec(x))]:}\begin{align*} \vec{\nabla} \cdot g(\vec{x}) & =-4 \pi G \int \mathrm{~d}^{3} x^{\prime} \rho\left(\vec{x}^{\prime}\right) \delta^{(3)}\left(\vec{x}-\vec{x}^{\prime}\right) \\ & =-4 \pi G \rho(\vec{x}) \tag{13.16} \end{align*}g(x)=4πG d3xρ(x)δ(3)(xx)(13.16)=4πGρ(x)
Since we also have g ( x ) = 2 Φ g ( x ) = 2 Φ vec(grad)*g( vec(x))=- vec(grad)^(2)Phi\vec{\nabla} \cdot g(\vec{x})=-\vec{\nabla}^{2} \Phig(x)=2Φ, we conclude that
(13.17) 2 Φ ( x ) = 4 π G ρ ( x ) (13.17) 2 Φ ( x ) = 4 π G ρ ( x ) {:(13.17) vec(grad)^(2)Phi( vec(x))=4pi G rho( vec(x)):}\begin{equation*} \vec{\nabla}^{2} \Phi(\vec{x})=4 \pi G \rho(\vec{x}) \tag{13.17} \end{equation*}(13.17)2Φ(x)=4πGρ(x)
This is the gravitational version of Poisson's equation. 9 9 ^(9){ }^{9}9
where g g vec(g)\vec{g}g is the gravitational field. The law can be expressed in terms of the potential Φ Φ Phi\PhiΦ (via g = Φ g = Φ vec(g)=- vec(grad)Phi\vec{g}=-\vec{\nabla} \Phig=Φ ). The potential field is the solution of Poisson's equation
9 9 ^(9){ }^{9}9 Recall that Laplace's equation for a field in the absence of sources says 2 Φ = 0 2 Φ = 0 vec(grad)^(2)Phi=0\vec{\nabla}^{2} \Phi=02Φ=0, but Poisson's equation for the field in the presence of sources says 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho2Φ=4πGρ, where 4 π G ρ 4 π G ρ 4pi G rho4 \pi G \rho4πGρ is the density of the source.
10 10 ^(10){ }^{10}10 Note that eqn 13.18 is the same as presented in eqn 0.12 , and eqn 13.19 is the same as presented in eqn 0.15 .
7 7 ^(7){ }^{7}7 Proof: Check by a direct computation that
(13.12) x ( 1 | x x | ) = x x | x x | 3 (13.12) x 1 x x = x x x x 3 {:(13.12) vec(grad)_( vec(x))((1)/(|( vec(x))- vec(x)^(')|))=-(( vec(x))- vec(x)^('))/(|( vec(x))- vec(x)^(')|^(3)):}\begin{equation*} \vec{\nabla}_{\vec{x}}\left(\frac{1}{\left|\vec{x}-\vec{x}^{\prime}\right|}\right)=-\frac{\vec{x}-\vec{x}^{\prime}}{\left|\vec{x}-\vec{x}^{\prime}\right|^{3}} \tag{13.12} \end{equation*}(13.12)x(1|xx|)=xx|xx|3
8 8 ^(8){ }^{8}8 See exercises for the proof that
(13.14) x 2 ( 1 | x x | ) = 4 π δ ( 3 ) ( x x ) (13.14) x 2 1 x x = 4 π δ ( 3 ) x x {:(13.14) vec(grad)_( vec(x))^(2)((1)/(|( vec(x))- vec(x)^(')|))=-4pidelta^((3))(( vec(x))- vec(x)^(')):}\begin{equation*} \vec{\nabla}_{\vec{x}}^{2}\left(\frac{1}{\left|\vec{x}-\vec{x}^{\prime}\right|}\right)=-4 \pi \delta^{(3)}\left(\vec{x}-\vec{x}^{\prime}\right) \tag{13.14} \end{equation*}(13.14)x2(1|xx|)=4πδ(3)(xx)
(13.19) 2 Φ ( x ) = 4 π G ρ ( x ) . (13.19) 2 Φ ( x ) = 4 π G ρ ( x ) . {:(13.19) vec(grad)^(2)Phi( vec(x))=4pi G rho( vec(x)).:}\begin{equation*} \vec{\nabla}^{2} \Phi(\vec{x})=4 \pi G \rho(\vec{x}) . \tag{13.19} \end{equation*}(13.19)2Φ(x)=4πGρ(x).
11 11 ^(11){ }^{11}11 The idea of the Green's function is that for a differential equation L ^ A ( x ) = L ^ A ( x ) = hat(L)A(x)=\hat{L} A(x)=L^A(x)= f ( x ) f ( x ) f(x)f(x)f(x), with L ^ L ^ hat(L)\hat{L}L^ representing a differential operator, we define the Green's function as L ^ G ( x , y ) = δ ( x y ) L ^ G ( x , y ) = δ ( x y ) hat(L)G(x,y)=delta(x-y)\hat{L} G(x, y)=\delta(x-y)L^G(x,y)=δ(xy). This is useful, as the Green's function can then be used to build a solution using the prescription
(13.20) A ( x ) = d y G ( x , y ) f ( y ) (13.20) A ( x ) = d y G ( x , y ) f ( y ) {:(13.20)A(x)=intdyG(x","y)f(y):}\begin{equation*} A(x)=\int \mathrm{d} y G(x, y) f(y) \tag{13.20} \end{equation*}(13.20)A(x)=dyG(x,y)f(y)
12 12 ^(12){ }^{12}12 We know that both the energy density and metric field are tensor fields, so we might expect Newton's law to be equivalent to one component of the Einstein equation.
Fig. 13.2 The relationship between the source of energy-momentum T T TTT and the metric field g g ggg in the Einstein equation. The source is constrained by conservation laws (via T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 ) affecting its coupling to the metric field that, in turn, determines the geometry. The metric field also determines the form of T T TTT itself.
13 13 ^(13){ }^{13}13 We can be a little more specific here. In electromagnetism (another linear theory), we often specify the currents and charges and then solve Maxwell's equations to find the electromagnetic fields. The same can be done in Newtonian gravity: specify the mass distribution and solve for Φ Φ Phi\PhiΦ. The same cannot be done in general relativity since the source of gravitation is the energy-momentum tensor T T T\boldsymbol{T}T, which is described in terms of the metric g g g\boldsymbol{g}g, precisely the field we are trying to find. This forces us to attempt to find consistent values of g g g\boldsymbol{g}g and T T T\boldsymbol{T}T simultaneously, which is very difficult.
Note the form of this equation that says that two derivatives of the potential field are proportional to the mass density ρ ρ rho\rhoρ, which is the source of the field. We therefore have a field theory of Newtonian gravitation that deals in a scalar field Φ ( x , t ) Φ ( x , t ) Phi( vec(x),t)\Phi(\vec{x}, t)Φ(x,t). We input a position in spacetime and output a function Φ Φ Phi\PhiΦ. This can be used to compute the gravitational field and hence the force on any mass.
Example 13.2
Green's functions are a very useful tool in solving equations like the Poisson equation. 11 11 ^(11){ }^{11}11 For our problem, we seek a solution to the equation
(13.21) 2 G ( x , x ) = 4 π δ ( 3 ) ( x x ) (13.21) 2 G x , x = 4 π δ ( 3 ) x x {:(13.21) vec(grad)^(2)G(( vec(x)), vec(x)^('))=4pidelta^((3))(( vec(x))- vec(x)^(')):}\begin{equation*} \vec{\nabla}^{2} G\left(\vec{x}, \vec{x}^{\prime}\right)=4 \pi \delta^{(3)}\left(\vec{x}-\vec{x}^{\prime}\right) \tag{13.21} \end{equation*}(13.21)2G(x,x)=4πδ(3)(xx)
We can then use the resulting Green's function G ( x , x ) G x , x G(( vec(x)), vec(x)^('))G\left(\vec{x}, \vec{x}^{\prime}\right)G(x,x) to build a solution to Poisson's equation using the prescription
(13.22) Φ ( x ) = G d 3 x G ( x , x ) ρ ( x ) (13.22) Φ ( x ) = G d 3 x G x , x ρ x {:(13.22)Phi( vec(x))=G intd^(3)x^(')G(( vec(x)), vec(x)^('))rho( vec(x)^(')):}\begin{equation*} \Phi(\vec{x})=G \int \mathrm{~d}^{3} x^{\prime} G\left(\vec{x}, \vec{x}^{\prime}\right) \rho\left(\vec{x}^{\prime}\right) \tag{13.22} \end{equation*}(13.22)Φ(x)=G d3xG(x,x)ρ(x)
Comparing to the steps above, we can identify the Green's function we need to solve Poisson's equation: it is simply
(13.23) G ( x , x ) = 1 | x x | (13.23) G x , x = 1 x x {:(13.23)G(( vec(x)), vec(x)^('))=-(1)/(|( vec(x))- vec(x)^(')|):}\begin{equation*} G\left(\vec{x}, \vec{x}^{\prime}\right)=-\frac{1}{\left|\vec{x}-\vec{x}^{\prime}\right|} \tag{13.23} \end{equation*}(13.23)G(x,x)=1|xx|
We shall try to fashion our theory of gravitation following the pattern set by Newton's field theory. In place of the Newtonian potential, we shall substitute the metric. In place of the mass density, we substitute the energy density. If we postulate that the Einstein equation has the same form as Poisson's equation for the Newtonian potential field, then Einstein's equation must look like
(13.24) ( 2 g ) = κ T (13.24) 2 g = κ T {:(13.24)(del^(2)g)=kappa T:}\begin{equation*} \left(\partial^{2} g\right)=\kappa T \tag{13.24} \end{equation*}(13.24)(2g)=κT
where κ κ kappa\kappaκ is a constant. 12 12 ^(12){ }^{12}12 That is to say that two derivatives of the metric field g g ggg are proportional to the energy density T T TTT.
Although these steps might make us optimistic, there must also be differences between Einstein and Newton's pictures of gravitation. First, Newton's law is instantaneous, running counter to the rules of special relativity. In addition, Fermat's principle says that light propagates in straight lines in flat spacetime but experiments show the bending of light from stars by gravitation, implying that gravity must warp spacetime. This means that, for Einstein's gravity, the geometry of curved space is caused by the energy density of matter which acts as a source (just like ρ ρ rho\rhoρ is a source of Φ ) Φ ) Phi)\Phi)Φ). However, the arrangement of energy density is itself determined by the geometry encoded in the metric field, leading to the situation shown in Fig. 13.2. We conclude that, in contrast to Newton, Einstein's gravitation is necessarily nonlinear: gravitational energy is acted on by gravity itself. 13 13 ^(13){ }^{13}13
We have now reached the point where we must fit the geometrical and physical fields together. The key physical principles determining the content on the right-hand side of a relativistic equation of motion are as follows. (i) Local causality: A signal can be sent between two points if, and only if, their separation is not spacelike. Colloquially: nothing travels faster than light. Our fields must therefore be consistent with special relativity. (ii) Local conservation of energy and momentum: All matter fields in the Universe have energy-momentum. This energy-momentum is locally conserved. That is to say, it must obey a continuity equation.
We also require energy to obey another condition. (iii) The dominant energy condition says that to any observer the local energy density appears non-negative and the local energy flow vector is not spacelike (i.e. it is timelike or null). We therefore demand that positive energy, which cannot move around a system at a speed faster than that of light, is locally conserved. 14 14 ^(14){ }^{14}14
All of the physical principles relating to the behaviour of mass energy can be embodied in an energy-momentum tensor field T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x), as described in the previous chapter. This tensor must describe positive energy and encodes local conservation of energy-momentum (at least approximately) via the constraint T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. The energy-momentum tensor vanishes only if all of the matter fields vanish. This means that all fields cost energy, and this is the source of gravitation.
In the next section, we shall, finally, attempt to marry the geometry of curvature embodied in R R R\boldsymbol{R}R and the mass energy of the fields of the Universe, expressed in T T T\boldsymbol{T}T in a tensor equation.

13.3 An incorrect guess

The apparently obvious field equation turns out to be incorrect.
The energy-momentum tensor has two slots or, equivalently, its components have two indices. We are therefore tempted to guess that the Einstein equation, that fixes the relationship of curvature (on the left) to the energy density of the matter fields (on the right) has a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) object on the left and T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν on the right. The obvious, but incorrect, natural candidate would be that the curvature-defining object is the Ricci tensor, with components R μ ν R μ ν R^(mu nu)R^{\mu \nu}Rμν. This would give us the field equation
(13.25) R μ ν = κ T μ ν (incorrect guess) (13.25) R μ ν = κ T μ ν  (incorrect guess)  {:(13.25)R^(mu nu)=kappaT^(mu nu)quad" (incorrect guess) ":}\begin{equation*} R^{\mu \nu}=\kappa T^{\mu \nu} \quad \text { (incorrect guess) } \tag{13.25} \end{equation*}(13.25)Rμν=κTμν (incorrect guess) 
with κ κ kappa\kappaκ a constant. 15 15 ^(15){ }^{15}15 But why is this wrong? The key is to remember that we must have the conservation equation T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0. We examine the consequence of this on our candidate field equation by taking the divergence
(13.26) R ; μ μ ν = κ T ; μ μ ν (incorrect guess). (13.26) R ; μ μ ν = κ T ; μ μ ν  (incorrect guess).  {:(13.26)R_(;mu)^(mu nu)=kappaT_(;mu)^(mu nu)quad" (incorrect guess). ":}\begin{equation*} R_{; \mu}^{\mu \nu}=\kappa T_{; \mu}^{\mu \nu} \quad \text { (incorrect guess). } \tag{13.26} \end{equation*}(13.26)R;μμν=κT;μμν (incorrect guess). 
Since T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 (or, equivalently T μ ν ; μ = 0 T μ ν ; μ = 0 T^(mu nu)_(;mu)=0T^{\mu \nu}{ }_{; \mu}=0Tμν;μ=0 ), the right-hand side of eqn 13.26 must vanish. What happens to the geometrical left-hand side?
14 14 ^(14){ }^{14}14 There is a less stringent weak energy condition, which says that the energy density measured by any observer must be non-negative.
15 This seems a good idea since the Ricci tensor is a sort of average of R R R\boldsymbol{R}R that retains those parts of curvature that, in a gravitational equation, can cause volumes to shrink. This was Einstein's original suggestion in October 1915.
16 16 ^(16){ }^{16}16 We make use of the symmetries of R α β γ δ R α β γ δ R_(alpha beta gamma delta)R_{\alpha \beta \gamma \delta}Rαβγδ, where swaps within the firs and second pairs of indices result in a minus sign.
17 17 ^(17){ }^{17}17 Do not confuse the Einstein tensor field G G G\boldsymbol{G}G with the gravitational constant G G GGG. It's unfortunate that they have the same symbol, but context should make clear which is which. In later chapters, we'll use units in which G = 1 G = 1 G=1G=1G=1.
18 18 ^(18){ }^{18}18 We use the compatibility condition g μ ν ; α = 0 g μ ν ; α = 0 g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0gμν;α=0 here.
19 19 ^(19){ }^{19}19 If you don't see this immediately, the steps are (i) raise the μ μ mu\muμ index in the second term to find R μ ν ; μ R α ν ; α R μ ν ; μ R α ν ; α R^(mu nu)_(;mu)-R^(alpha nu)_(;alpha)R^{\mu \nu}{ }_{; \mu}-R^{\alpha \nu}{ }_{; \alpha}Rμν;μRαν;α; (ii) rename the α α alpha\alphaα index μ μ mu\muμ and the expression vanishes.
20 20 ^(20){ }^{20}20 Remember that the effect of the covariant derivative on a scalar field is simply
f = f ; μ = f , μ = f x μ . f = f ; μ = f , μ = f x μ . grad f=f_(;mu)=f_(,mu)=(del f)/(delx^(mu)).\nabla f=f_{; \mu}=f_{, \mu}=\frac{\partial f}{\partial x^{\mu}} .f=f;μ=f,μ=fxμ.
To answer this, we need a mathematical result to constrain our equations.

Example 13.3

As we'll discuss in Chapter 43, the components of the Riemann tensor satisfy a geometrical constraint known as the Bianchi identity, whose component form is written as
(13.27) R β γ δ ; λ α + R β λ γ ; δ α + R β δ λ ; γ α = 0 . (13.27) R β γ δ ; λ α + R β λ γ ; δ α + R β δ λ ; γ α = 0 . {:(13.27)R_(beta gamma delta;lambda)^(alpha)+R_(beta lambda gamma;delta)^(alpha)+R_(beta delta lambda;gamma)^(alpha)=0.:}\begin{equation*} R_{\beta \gamma \delta ; \lambda}^{\alpha}+R_{\beta \lambda \gamma ; \delta}^{\alpha}+R_{\beta \delta \lambda ; \gamma}^{\alpha}=0 . \tag{13.27} \end{equation*}(13.27)Rβγδ;λα+Rβλγ;δα+Rβδλ;γα=0.
For now, it is enough to say that this identity comes from a general principle that 'the boundary of a boundary is zero', an idea that leads to the well-known vector identity that the divergence of the curl of a vector field vanishes. We can contract indices in this identity, setting α = λ α = λ alpha=lambda\alpha=\lambdaα=λ, to find 16 16 ^(16){ }^{16}16
(13.28) R β γ δ ; α α + R β γ ; δ R β δ ; γ = 0 (13.28) R β γ δ ; α α + R β γ ; δ R β δ ; γ = 0 {:(13.28)R_(beta gamma delta;alpha)^(alpha)+R_(beta gamma;delta)-R_(beta delta;gamma)=0:}\begin{equation*} R_{\beta \gamma \delta ; \alpha}^{\alpha}+R_{\beta \gamma ; \delta}-R_{\beta \delta ; \gamma}=0 \tag{13.28} \end{equation*}(13.28)Rβγδ;αα+Rβγ;δRβδ;γ=0
Now raise the β β beta\betaβ index and contract with the δ δ delta\deltaδ index to get
R γ ; α α + R γ ; β β R ; γ = 0 (13.29) 2 R α γ ; α = R ; γ . R γ ; α α + R γ ; β β R ; γ = 0 (13.29) 2 R α γ ; α = R ; γ . {:[R_(gamma;alpha)^(alpha)+R_(gamma;beta)^(beta)-R_(;gamma)=0],[(13.29)2R^(alpha)_(gamma;alpha)=R_(;gamma).]:}\begin{align*} R_{\gamma ; \alpha}^{\alpha}+R_{\gamma ; \beta}^{\beta}-R_{; \gamma} & =0 \\ 2 R^{\alpha}{ }_{\gamma ; \alpha} & =R_{; \gamma} . \tag{13.29} \end{align*}Rγ;αα+Rγ;ββR;γ=0(13.29)2Rαγ;α=R;γ.
This equation is known as the contracted Bianchi identity and we will use it below.
To examine the consequence of the identity discussed in the last example, we now define the trace-reversed Ricci tensor, also known as the Einstein tensor field, as the ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor G ( x ) G ( x ) G(x)\boldsymbol{G}(x)G(x) with components 17 17 ^(17){ }^{17}17
(13.30) G μ ν ( x ) = R μ ν ( x ) 1 2 g μ ν ( x ) R ( x ) (13.30) G μ ν ( x ) = R μ ν ( x ) 1 2 g μ ν ( x ) R ( x ) {:(13.30)G^(mu nu)(x)=R^(mu nu)(x)-(1)/(2)g^(mu nu)(x)R(x):}\begin{equation*} G^{\mu \nu}(x)=R^{\mu \nu}(x)-\frac{1}{2} g^{\mu \nu}(x) R(x) \tag{13.30} \end{equation*}(13.30)Gμν(x)=Rμν(x)12gμν(x)R(x)
Taking the divergence of the Einstein tensor G G grad*G\boldsymbol{\nabla} \cdot \boldsymbol{G}G, we find that 18 18 ^(18){ }^{18}18
(13.31) G μ ν ; μ = R ; μ μ ν 1 2 g μ ν R ; μ (13.31) G μ ν ; μ = R ; μ μ ν 1 2 g μ ν R ; μ {:(13.31)G^(mu nu)_(;mu)=R_(;mu)^(mu nu)-(1)/(2)g^(mu nu)R_(;mu):}\begin{equation*} G^{\mu \nu}{ }_{; \mu}=R_{; \mu}^{\mu \nu}-\frac{1}{2} g^{\mu \nu} R_{; \mu} \tag{13.31} \end{equation*}(13.31)Gμν;μ=R;μμν12gμνR;μ
Using the contracted Bianchi identity ( R ; μ = 2 R α μ ; α R ; μ = 2 R α μ ; α R_(;mu)=2R^(alpha)_(mu;alpha)R_{; \mu}=2 R^{\alpha}{ }_{\mu ; \alpha}R;μ=2Rαμ;α ) on the second term on the right, we find 19 19 ^(19){ }^{19}19
(13.32) G ; μ μ ν = R ; μ μ ν g μ ν R μ ; α α = 0 (13.32) G ; μ μ ν = R ; μ μ ν g μ ν R μ ; α α = 0 {:(13.32)G_(;mu)^(mu nu)=R_(;mu)^(mu nu)-g^(mu nu)R_(mu;alpha)^(alpha)=0:}\begin{equation*} G_{; \mu}^{\mu \nu}=R_{; \mu}^{\mu \nu}-g^{\mu \nu} R_{\mu ; \alpha}^{\alpha}=0 \tag{13.32} \end{equation*}(13.32)G;μμν=R;μμνgμνRμ;αα=0
We conclude that the Einstein tensor has the property
(13.33) G = ( R μ ν 1 2 g μ ν R ) ; μ = 0 (13.33) G = R μ ν 1 2 g μ ν R ; μ = 0 {:(13.33)grad*G=(R^(mu nu)-(1)/(2)g^(mu nu)R)_(;mu)=0:}\begin{equation*} \boldsymbol{\nabla} \cdot \boldsymbol{G}=\left(R^{\mu \nu}-\frac{1}{2} g^{\mu \nu} R\right)_{; \mu}=0 \tag{13.33} \end{equation*}(13.33)G=(Rμν12gμνR);μ=0
Although this doesn't look particularly scandalous, it causes a major problem for the divergence of our candidate equation: R μ ν ; μ = κ T μ ν ; μ = R μ ν ; μ = κ T μ ν ; μ = R^(mu nu)_(;mu)=kappaT^(mu nu)_(;mu)=R^{\mu \nu}{ }_{; \mu}=\kappa T^{\mu \nu}{ }_{; \mu}=Rμν;μ=κTμν;μ= 0 . We combine this expression with eqn 13.33 to conclude
(13.34) 1 2 g μ ν R ; μ = 1 2 R ; ν = 0 (13.34) 1 2 g μ ν R ; μ = 1 2 R ; ν = 0 {:(13.34)-(1)/(2)*g^(mu nu)R_(;mu)=-(1)/(2)R^(;nu)=0:}\begin{equation*} -\frac{1}{2} \cdot g^{\mu \nu} R_{; \mu}=-\frac{1}{2} R^{; \nu}=0 \tag{13.34} \end{equation*}(13.34)12gμνR;μ=12R;ν=0
The guts of this latter equation 20 ( R ; ν = 0 ) 20 R ; ν = 0 ^(20)(R^(;nu)=0){ }^{20}\left(R^{; \nu}=0\right)20(R;ν=0) forces the Ricci scalar field R ( x ) R ( x ) R(x)R(x)R(x) to be a constant throughout the Universe. This on its own is not
a problem. However, the trouble comes when we examine the candiate equation (eqn 13.25) again, since we find that
(13.35) R ( x ) = g μ ν ( x ) R μ ν ( x ) = κ g μ ν ( x ) T μ ν ( x ) = const. (13.35) R ( x ) = g μ ν ( x ) R μ ν ( x ) = κ g μ ν ( x ) T μ ν ( x ) =  const.  {:(13.35)R(x)=g^(mu nu)(x)R_(mu nu)(x)=kappag^(mu nu)(x)T_(mu nu)(x)=" const. ":}\begin{equation*} R(x)=g^{\mu \nu}(x) R_{\mu \nu}(x)=\kappa g^{\mu \nu}(x) T_{\mu \nu}(x)=\text { const. } \tag{13.35} \end{equation*}(13.35)R(x)=gμν(x)Rμν(x)=κgμν(x)Tμν(x)= const. 
That is, the trace g μ ν T μ ν g μ ν T μ ν g^(mu nu)T_(mu nu)g^{\mu \nu} T_{\mu \nu}gμνTμν of T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν must be constant throughout the Universe. This is unreasonable! It implies a uniform distribution of matter and energy throughout the Universe which is at odds with our experience, where matter is certainly non-uniform. We are forced to drop the candidate equation.

13.4 Einstein's field equation

The problem encountered in the last section leads to the solution.
On our way to proving that our first attempt at a field equation was inadequate, we identified a new ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) [or ( 0 , 2 ) ] ( 0 , 2 ) ] (0,2)](0,2)](0,2)] tensor G G G\boldsymbol{G}G, which had components G μ ν = R μ ν 1 2 g μ ν R G μ ν = R μ ν 1 2 g μ ν R G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)RG_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} RGμν=Rμν12gμνR. It is this object that provides the solution to our problem. Therefore, having discarded our first guess, we now consider instead a new candidate field equation 21 21 ^(21){ }^{21}21
(13.36) G μ ν = κ T μ ν (13.36) G μ ν = κ T μ ν {:(13.36)G_(mu nu)=kappaT_(mu nu):}\begin{equation*} G_{\mu \nu}=\kappa T_{\mu \nu} \tag{13.36} \end{equation*}(13.36)Gμν=κTμν
We know from the last section that the divergence of this equation is zero on both sides. This, it turns out, is exactly the field equation for which we've been searching. However, we have yet to work out the constant of proportionality κ κ kappa\kappaκ. This can be evaluated using the condition that this equation is compatible with the results of Newtonian gravitation in the limit of weak fields (i.e. with the equation 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho2Φ=4πGρ ).
Example 13.4
In the weak-field limit, spacetime is only very slightly curved and therefore the metric is very close to the flat Minkowski metric, so g 00 g 00 1 g 00 g 00 1 g_(00)~~g^(00)~~-1g_{00} \approx g^{00} \approx-1g00g001 and g i i g i i 1 g i i g i i 1 g_(ii)~~g^(ii)~~1g_{i i} \approx g^{i i} \approx 1giigii1. As a result, we have for any tensor A A A\boldsymbol{A}A in this geometry, that A 00 A 00 A 00 A 00 A^(00)~~A_(00)A^{00} \approx A_{00}A00A00, along with A i i = A i i A i i = A i i A^(ii)=A_(ii)A^{i i}=A_{i i}Aii=Aii. Moreover, the energy-momentum tensor is dominated by the energy density part (because ρ p / c 2 ρ p / c 2 rho≫p//c^(2)\rho \gg p / c^{2}ρp/c2 ) so the only significant element of T T T\boldsymbol{T}T is T 00 = ρ c 2 T 00 = ρ c 2 T^(00)=rhoc^(2)T^{00}=\rho c^{2}T00=ρc2, with all other elements vanishingly small. This means that, for all of the spatial (ij) components of the Einstein tensor, we have
(13.37) G i j = R i j 1 2 g i j R = 0 (13.37) G i j = R i j 1 2 g i j R = 0 {:(13.37)G^(ij)=R^(ij)-(1)/(2)g^(ij)R=0:}\begin{equation*} G^{i j}=R^{i j}-\frac{1}{2} g^{i j} R=0 \tag{13.37} \end{equation*}(13.37)Gij=Rij12gijR=0
from which we find R 11 = R 22 = R 33 = R / 2 R 11 = R 22 = R 33 = R / 2 R^(11)=R^(22)=R^(33)=R//2R^{11}=R^{22}=R^{33}=R / 2R11=R22=R33=R/2 and so
(13.38) R 11 + R 22 + R 33 = 3 R / 2 (13.38) R 11 + R 22 + R 33 = 3 R / 2 {:(13.38)R^(11)+R^(22)+R^(33)=3R//2:}\begin{equation*} R^{11}+R^{22}+R^{33}=3 R / 2 \tag{13.38} \end{equation*}(13.38)R11+R22+R33=3R/2
As a result, R = g μ ν R μ ν = R 00 + 3 2 R R = g μ ν R μ ν = R 00 + 3 2 R R=g_(mu nu)R^(mu nu)=-R^(00)+(3)/(2)RR=g_{\mu \nu} R^{\mu \nu}=-R^{00}+\frac{3}{2} RR=gμνRμν=R00+32R and therefore R = 2 R 00 R = 2 R 00 R=2R^(00)R=2 R^{00}R=2R00. This allows us to calculate G 00 = R 00 1 2 g 00 R G 00 = R 00 1 2 g 00 R G^(00)=R^(00)-(1)/(2)g^(00)RG^{00}=R^{00}-\frac{1}{2} g^{00} RG00=R0012g00R which gives
(13.39) G 00 = 2 R 00 2 R 00 (13.39) G 00 = 2 R 00 2 R 00 {:(13.39)G^(00)=2R^(00)~~2R_(00):}\begin{equation*} G^{00}=2 R^{00} \approx 2 R_{00} \tag{13.39} \end{equation*}(13.39)G00=2R002R00
Recall that, with the weak-field metric, the only appreciable connection coefficients are (replacing factors of c) Γ 00 i = ( Φ / c 2 ) / x i Γ 00 i = Φ / c 2 / x i Gamma_(00)^(i)=del(Phi//c^(2))//delx^(i)\Gamma_{00}^{i}=\partial\left(\Phi / c^{2}\right) / \partial x^{i}Γ00i=(Φ/c2)/xi and so R i 0 j 0 = 2 ( Φ / c 2 ) / x i x j R i 0 j 0 = 2 Φ / c 2 / x i x j R^(i)_(0j0)=del^(2)(Phi//c^(2))//delx^(i)delx^(j)R^{i}{ }_{0 j 0}=\partial^{2}\left(\Phi / c^{2}\right) / \partial x^{i} \partial x^{j}Ri0j0=2(Φ/c2)/xixj. We can then read off 22 22 ^(22){ }^{22}22 that R 00 = ( 1 / c 2 ) 2 Φ R 00 = 1 / c 2 2 Φ R_(00)=(1//c^(2)) vec(grad)^(2)PhiR_{00}=\left(1 / c^{2}\right) \vec{\nabla}^{2} \PhiR00=(1/c2)2Φ and so
(13.40) G 00 = 2 c 2 2 Φ . (13.40) G 00 = 2 c 2 2 Φ . {:(13.40)G^(00)=(2)/(c^(2))* vec(grad)^(2)Phi.:}\begin{equation*} G^{00}=\frac{2}{c^{2}} \cdot \vec{\nabla}^{2} \Phi . \tag{13.40} \end{equation*}(13.40)G00=2c22Φ.
21 21 ^(21){ }^{21}21 In case this mathematics is a distraction, here's a physics recap. The Ricci tensor represents an average of the Riemann curvature tensor that will encode those parts of curvature that cause volumes to shrink owing to the gravitational interaction. However, equating the Ricci tensor to the energying the Ricci tensor to the energy-
momentum tensor gives a theory that momentum tensor gives a theory that
doesn't conserve energy (i.e. it is incomdoesn't conserve energy (i.e. it is incom-
patible with T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 ). The Bianchi identity (geometrically) encodes the desired energy conservation. Using a trace-reversed Ricci tensor, known as the Einstein tensor, builds energy conservation back into the theory since, via the Bianchi identity, it allows the resulting equations to conserve energy.
In this example, we will temporarily reinsert the factors of c c ccc so it's a bit more clear how large various factors are.
22 Explicitly we have R 00 = i = 1 3 R 0 i 0 i . 22  Explicitly we have  R 00 = i = 1 3 R 0 i 0 i {:[^(22)" Explicitly we have "R_(00)=],[sum_(i=1)^(3)R_(0i0)^(i)". "]:}\begin{aligned} & { }^{22} \text { Explicitly we have } R_{00}= \\ & \sum_{i=1}^{3} R_{0 i 0}^{i} \text {. } \end{aligned}22 Explicitly we have R00=i=13R0i0i
Putting this together with T 00 = ρ c 2 T 00 = ρ c 2 T^(00)=rhoc^(2)T^{00}=\rho c^{2}T00=ρc2, the Einstein equation G = κ T G = κ T G=kappa T\boldsymbol{G}=\kappa \boldsymbol{T}G=κT predicts a universal field equation of
(13.41) 2 Φ = κ c 4 2 ρ . (13.41) 2 Φ = κ c 4 2 ρ . {:(13.41) vec(grad)^(2)Phi=(kappac^(4))/(2)rho.:}\begin{equation*} \vec{\nabla}^{2} \Phi=\frac{\kappa c^{4}}{2} \rho . \tag{13.41} \end{equation*}(13.41)2Φ=κc42ρ.
This can be compared with Poisson's equation for gravitation (the field equation that corresponds to Newton's universal law of gravitation), which reads 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ vec(grad)^(2)Phi=4pi G rho\vec{\nabla}^{2} \Phi=4 \pi G \rho2Φ=4πGρ. Hence, we deduce that
(13.42) κ = 8 π G c 4 (13.42) κ = 8 π G c 4 {:(13.42)kappa=(8pi G)/(c^(4)):}\begin{equation*} \kappa=\frac{8 \pi G}{c^{4}} \tag{13.42} \end{equation*}(13.42)κ=8πGc4
In SI units, the factor is κ = 8 π G / c 4 2.08 × 10 43 N 1 κ = 8 π G / c 4 2.08 × 10 43 N 1 kappa=8pi G//c^(4)~~2.08 xx10^(-43)N^(-1)\kappa=8 \pi G / c^{4} \approx 2.08 \times 10^{-43} \mathrm{~N}^{-1}κ=8πG/c42.08×1043 N1.
The result of the previous example, setting c = 1 c = 1 c=1c=1c=1 again, is that κ = 8 π G κ = 8 π G kappa=8pi G\kappa=8 \pi Gκ=8πG. We are now in a position to write a field equation for gravity, known as
23 23 ^(23){ }^{23}23 This is the equation Einstein presented in November 1915, after having rejected his first attempt.
the Einstein field equation. 23 23 ^(23){ }^{23}23
The Einstein field equation, version 1
In coordinates,
(13.43) ( R μ ν 1 2 g μ ν R ) = 8 π G T μ ν (13.43) R μ ν 1 2 g μ ν R = 8 π G T μ ν {:(13.43)(R_(mu nu)-(1)/(2)g_(mu nu)R)=8pi GT_(mu nu):}\begin{equation*} \left(R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R\right)=8 \pi G T_{\mu \nu} \tag{13.43} \end{equation*}(13.43)(Rμν12gμνR)=8πGTμν
In coordinate-free notation, this becomes
(13.44) G ( x ) = 8 π G T ( x ) (13.44) G ( x ) = 8 π G T ( x ) {:(13.44)G(x)=8pi GT(x):}\begin{equation*} \boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x) \tag{13.44} \end{equation*}(13.44)G(x)=8πGT(x)

Example 13.5

As a further manipulation, let's take the trace of eqn 13.43. This is achieved by multiplying by g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν, so that we get
(11.45) g μ ν R μ ν 1 2 g μ ν g μ ν R = 8 π G g μ ν T μ ν (11.45) g μ ν R μ ν 1 2 g μ ν g μ ν R = 8 π G g μ ν T μ ν {:(11.45)g^(mu nu)R_(mu nu)-(1)/(2)*g^(mu nu)g_(mu nu)R=8pi Gg^(mu nu)T_(mu nu):}\begin{equation*} g^{\mu \nu} R_{\mu \nu}-\frac{1}{2} \cdot g^{\mu \nu} g_{\mu \nu} R=8 \pi G g^{\mu \nu} T_{\mu \nu} \tag{11.45} \end{equation*}(11.45)gμνRμν12gμνgμνR=8πGgμνTμν
Let's now consider the trace T T TTT of the energy-momentum tensor T = g μ ν T μ ν T = g μ ν T μ ν T=g^(mu nu)T_(mu nu)T=g^{\mu \nu} T_{\mu \nu}T=gμνTμν, and hence eqn 13.45 gives (using g μ ν g μ ν = 4 g μ ν g μ ν = 4 g^(mu nu)g_(mu nu)=4g^{\mu \nu} g_{\mu \nu}=4gμνgμν=4 and R = g μ ν R μ ν R = g μ ν R μ ν R=g_(mu nu)R^(mu nu)R=g_{\mu \nu} R^{\mu \nu}R=gμνRμν ) R 2 R = R = 8 π G T R 2 R = R = 8 π G T R-2R=-R=8pi GTR-2 R=-R=8 \pi G TR2R=R=8πGT This implies we can also write
(13.46) R μ ν + 1 2 g μ ν ( 8 π G T ) = 8 π G T μ ν (13.46) R μ ν + 1 2 g μ ν ( 8 π G T ) = 8 π G T μ ν {:(13.46)R_(mu nu)+(1)/(2)g_(mu nu)(8pi GT)=8pi GT_(mu nu):}\begin{equation*} R_{\mu \nu}+\frac{1}{2} g_{\mu \nu}(8 \pi G T)=8 \pi G T_{\mu \nu} \tag{13.46} \end{equation*}(13.46)Rμν+12gμν(8πGT)=8πGTμν
The previous example therefore leads to a second form for the Einstein field equation:

The Einstein field equation, version 2

(13.47) R μ ν = 8 π G ( T μ ν 1 2 g μ ν T ) (13.47) R μ ν = 8 π G T μ ν 1 2 g μ ν T {:(13.47)R_(mu nu)=8pi G(T_(mu nu)-(1)/(2)g_(mu nu)T):}\begin{equation*} R_{\mu \nu}=8 \pi G\left(T_{\mu \nu}-\frac{1}{2} g_{\mu \nu} T\right) \tag{13.47} \end{equation*}(13.47)Rμν=8πG(Tμν12gμνT)
Our method of arriving at the Einstein equation has been rather haphazard. We might ask if we can add other terms to the right-hand side of the Einstein equation. In fact, it is permissible to add any scalar multiple of the metric field tensor, so we can take our (version 1) equation G ( x ) = 8 π G T ( x ) G ( x ) = 8 π G T ( x ) G(x)=8pi GT(x)\boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x)G(x)=8πGT(x) and put an extra Λ g ( x ) Λ g ( x ) -Lambda g(x)-\Lambda \boldsymbol{g}(x)Λg(x) on the right-hand side and if Λ Λ Lambda\LambdaΛ is just a constant then our new equation will still work. In fact, the property that α g = g μ ν ; α = 0 α g = g μ ν ; α = 0 grad_(alpha)g=g_(mu nu;alpha)=0\nabla_{\alpha} \boldsymbol{g}=g_{\mu \nu ; \alpha}=0αg=gμν;α=0 means that the divergence arguments
we have been using won't be troubled by this extra term. We therefore write our third version of the Einstein equation as follows:

The Einstein field equation, version 3

(13.48) ( R μ ν 1 2 g μ ν R ) = 8 π G T μ ν Λ g μ ν , (13.49) G ( x ) = 8 π G T ( x ) Λ g ( x ) . (13.48) R μ ν 1 2 g μ ν R = 8 π G T μ ν Λ g μ ν , (13.49) G ( x ) = 8 π G T ( x ) Λ g ( x ) . {:[(13.48)(R_(mu nu)-(1)/(2)g_(mu nu)R)=8pi GT_(mu nu)-Lambdag_(mu nu)","],[(13.49)G(x)=8pi GT(x)-Lambda g(x).]:}\begin{gather*} \left(R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R\right)=8 \pi G T_{\mu \nu}-\Lambda g_{\mu \nu}, \tag{13.48}\\ \boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x)-\Lambda \boldsymbol{g}(x) . \tag{13.49} \end{gather*}(13.48)(Rμν12gμνR)=8πGTμνΛgμν,(13.49)G(x)=8πGT(x)Λg(x).
The constant of proportionality Λ Λ Lambda\LambdaΛ is known as the cosmological constant. Physically, the cosmological constant represents an extra source of energy in the Universe, in addition to the matter fields. Owing to the difference in sign compared to positive T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν (as demanded by the dominant energy condition), a positive value of Λ Λ Lambda\LambdaΛ will give a repulsive gravitational interaction.
In order to retain the simplicity of the equation G = 8 π G T G = 8 π G T G=8pi GT\boldsymbol{G}=8 \pi G \boldsymbol{T}G=8πGT, we could treat the cosmological constant term as if it were a source of energymomentum and write
(13.50) G = 8 π G ( T mat + T vac ) (13.50) G = 8 π G T mat + T vac {:(13.50)G=8pi G(T^(mat)+T^(vac)):}\begin{equation*} \boldsymbol{G}=8 \pi G\left(\boldsymbol{T}^{\mathrm{mat}}+\boldsymbol{T}^{\mathrm{vac}}\right) \tag{13.50} \end{equation*}(13.50)G=8πG(Tmat+Tvac)
where T mat T mat  T^("mat ")\boldsymbol{T}^{\text {mat }}Tmat  is the energy-momentum tensor for matter that we have described in this chapter and
(13.51) T vac = Λ 8 π G g (13.51) T vac = Λ 8 π G g {:(13.51)T^(vac)=-(Lambda)/(8pi G)g:}\begin{equation*} \boldsymbol{T}^{\mathrm{vac}}=-\frac{\Lambda}{8 \pi G} \boldsymbol{g} \tag{13.51} \end{equation*}(13.51)Tvac=Λ8πGg
is a contribution to the energy-momentum tensor from sources other than the matter fields. The introduction and rejection of Λ Λ Lambda\LambdaΛ by Einstein has become a famous story in physics folklore. We shall discuss its consequences in Part III of the book. 24 24 ^(24){ }^{24}24
We have arrived at an equation of motion for the metric field that determines gravitation. Our task is now to explore its consequences on our Universe. In the coming chapters, it is useful to keep in mind a few helpful heuristics we have seen already in dealing with general relativity.
A metric is visualized via its light cones. These are found using d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 along null geodesics.
  • We use two types of reference frame: (i) a coordinate frame, where our geometric arguments are often simplest; (ii) an orthonormal frame, where observers make measurements.
  • Coordinates have no intrinsic metric significance but, in cases of spherical symmetry, useful coordinates to have in mind are spherical polars ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ).
  • Indices are raised/lowered in the coordinate frame using the metric. Indices in orthonormal frames are manipulated with the Minkowski tensor. The connection coefficients don't vanish in the orthonormal frame. If we want this we need to shift to a local inertial frame.
    24 24 ^(24){ }^{24}24 A brief history of solutions involving
    Λ Λ Lambda\LambdaΛ is given in Chapter 18.
  • General covariance says that a valid tensor equation in flat space is a valid tensor equation in curved space. To upgrade tensor equations, exchange η g η g eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}ηg. To upgrade derivatives, use the comma goes to semicolon rule.
  • When manipulating vectors, it's often helpful to use (i) g μ ν = g μ ν = g_(mu nu)=g_{\mu \nu}=gμν= e μ e ν e μ e ν e_(mu)*e_(nu)\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}eμeν; (ii) g μ ν ; α = 0 g μ ν ; α = 0 g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0gμν;α=0. For timelike velocities u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1.
  • The perfect fluid, in the orthonormal frame for spherical polars, has components
T t ^ t ^ = ρ , T r ^ r ^ = p , T θ ^ θ ^ = p , T ϕ ^ ϕ ^ = p T t ^ t ^ = ρ , T r ^ r ^ = p , T θ ^ θ ^ = p , T ϕ ^ ϕ ^ = p T_( hat(t) hat(t))=rho,quadT_( hat(r) hat(r))=p,quadT_( hat(theta) hat(theta))=p,quadT_( hat(phi) hat(phi))=pT_{\hat{t} \hat{t}}=\rho, \quad T_{\hat{r} \hat{r}}=p, \quad T_{\hat{\theta} \hat{\theta}}=p, \quad T_{\hat{\phi} \hat{\phi}}=pTt^t^=ρ,Tr^r^=p,Tθ^θ^=p,Tϕ^ϕ^=p

Chapter summary

  • The left-hand side of the Einstein field equation encodes geometry using the Ricci tensor and scalar to form the Einstein tensor G G G\boldsymbol{G}G with components
(13.52) G μ ν = R μ ν 1 2 g μ ν R (13.52) G μ ν = R μ ν 1 2 g μ ν R {:(13.52)G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)R:}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R \tag{13.52} \end{equation*}(13.52)Gμν=Rμν12gμνR
  • The right-hand side of Einstein's equation encodes the energymomentum of matter fields using the tensor T T T\boldsymbol{T}T.
  • These are tied together in Einstein's equation, subject to the key constraint that energy-momentum is locally conserved. The Einstein equation is G ( x ) = 8 π G T ( x ) Λ g ( x ) G ( x ) = 8 π G T ( x ) Λ g ( x ) G(x)=8pi GT(x)-Lambda g(x)\boldsymbol{G}(x)=8 \pi G \boldsymbol{T}(x)-\Lambda \boldsymbol{g}(x)G(x)=8πGT(x)Λg(x) and has components
(13.53) ( R μ ν 1 2 g μ ν R ) = 8 π G T μ ν Λ g μ ν (13.53) R μ ν 1 2 g μ ν R = 8 π G T μ ν Λ g μ ν {:(13.53)(R_(mu nu)-(1)/(2)g_(mu nu)R)=8pi GT_(mu nu)-Lambdag_(mu nu):}\begin{equation*} \left(R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R\right)=8 \pi G T_{\mu \nu}-\Lambda g_{\mu \nu} \tag{13.53} \end{equation*}(13.53)(Rμν12gμνR)=8πGTμνΛgμν

Exercises

(13.1) Show that Newton's second law and his law of gravitation are invariant with respect to Galilean transformations.
(13.2) Consider the equation
(13.54) 2 ( 1 r ) = 4 π δ ( x ) (13.54) 2 1 r = 4 π δ ( x ) {:(13.54) vec(grad)^(2)((1)/(r))=-4pi delta( vec(x)):}\vec{\nabla}^{2}\left(\begin{array}{l} \frac{1}{r} \tag{13.54} \end{array}\right)=-4 \pi \delta(\vec{x})(13.54)2(1r)=4πδ(x)
where r = | x | r = | x | r=| vec(x)|r=|\vec{x}|r=|x|. We shall prove this. (a) Evaluate x 2 r x 2 r (del)/(delx^(2))r\frac{\partial}{\partial x^{2}} rx2r and x 2 r 1 x 2 r 1 (del)/(delx^(2))r^(-1)\frac{\partial}{\partial x^{2}} r^{-1}x2r1. (b) Show that
(13.55) 2 x i x j ( 1 r ) = ( 3 n i n j δ i j ) r 3 (13.55) 2 x i x j 1 r = 3 n i n j δ i j r 3 {:(13.55)(del^(2))/(delx^(i)delx^(j))((1)/(r))=((3n^(i)n^(j)-delta^(ij)))/(r^(3)):}\begin{equation*} \frac{\partial^{2}}{\partial x^{i} \partial x^{j}}\left(\frac{1}{r}\right)=\frac{\left(3 n^{i} n^{j}-\delta^{i j}\right)}{r^{3}} \tag{13.55} \end{equation*}(13.55)2xixj(1r)=(3ninjδij)r3
where n i = x i / r n i = x i / r n^(i)=x^(i)//rn^{i}=x^{i} / rni=xi/r. (c) Use the previous result to show that 2 ( r 1 ) = 0 2 r 1 = 0 vec(grad)^(2)(r^(-1))=0\vec{\nabla}^{2}\left(r^{-1}\right)=02(r1)=0 for | x | 0 | x | 0 | vec(x)|!=0|\vec{x}| \neq 0|x|0. (d) Defining j = ( r 1 ) j = r 1 vec(j)= vec(grad)(r^(-1))\vec{j}=\vec{\nabla}\left(r^{-1}\right)j=(r1), use Gauss' theorem to show that
(13.56) V d 3 x j = 4 π (13.56) V d 3 x j = 4 π {:(13.56)int_(V)d^(3)x vec(grad)* vec(j)=-4pi:}\begin{equation*} \int_{V} \mathrm{~d}^{3} x \vec{\nabla} \cdot \vec{j}=-4 \pi \tag{13.56} \end{equation*}(13.56)V d3xj=4π
where V V VVV is a volume that encloses x = 0 x = 0 vec(x)=0\vec{x}=0x=0. (e) Use the previous result to complete the proof.
(13.3) Verify that taking the trace Tr ( G μ ν ) Tr G μ ν Tr(G^(mu nu))\operatorname{Tr}\left(G^{\mu \nu}\right)Tr(Gμν) of the Einstein tensor with components G μ ν G μ ν G^(mu nu)G^{\mu \nu}Gμν gives Tr ( G μ ν ) = R Tr G μ ν = R Tr(G^(mu nu))=-R\operatorname{Tr}\left(G^{\mu \nu}\right)=-RTr(Gμν)=R, where R R RRR is the trace of the Ricci tensor.